Department of Mathematics & Statistics
MATH 160/101
2016
Algebra Outline Notes
v
z u y x
Contents 0 Course Information for MATH160/101/102
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1 Vectors 1.1 Introduction . . . . . . . . . . . . . . . . . . . . 1.2 2- and 3-dimensional vectors . . . . . . . . . . . 1.3 Parallelograms . . . . . . . . . . . . . . . . . . . 1.4 Scalar multiplication . . . . . . . . . . . . . . . 1.5 The magnitude . . . . . . . . . . . . . . . . . . 1.6 Unit vectors . . . . . . . . . . . . . . . . . . . . 1.7 Parallel vectors . . . . . . . . . . . . . . . . . . 1.8 Applications to geometry . . . . . . . . . . . . . 1.9 The scalar product . . . . . . . . . . . . . . . . 1.10 The parametric equations of a line . . . . . . . . 1.11 Projections . . . . . . . . . . . . . . . . . . . . 1.12 Distance from a point to a line . . . . . . . . . . 1.13 The normal equation of a plane . . . . . . . . . 1.14 Angle between two planes . . . . . . . . . . . . 1.15 Intersection of a line and a plane . . . . . . . . 1.16 Distance from a point to a plane . . . . . . . . . 1.17 Projections onto a plane . . . . . . . . . . . . . 1.18 The parametric equation of a plane . . . . . . . 1.19 An application of vectors to calculus . . . . . . 1.20 An application of vectors to computer graphics .
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1 1 4 5 6 8 9 10 11 13 17 20 21 22 24 25 26 27 29 31 33
2 Linear equations and matrices 2.1 Systems of linear equations . . . . . . . . . 2.2 Augmented matrices . . . . . . . . . . . . 2.3 Operations on equations . . . . . . . . . . 2.4 Equivalent systems . . . . . . . . . . . . . 2.5 Gauss reduction . . . . . . . . . . . . . . . 2.6 Multiple solutions and inconsistent systems 2.7 Matrices . . . . . . . . . . . . . . . . . . . 2.8 Matrix operations . . . . . . . . . . . . . . 2.9 Identity and inverse matrices . . . . . . . . 2.10 Applications of matrices . . . . . . . . . .
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35 35 36 37 38 39 45 48 50 55 57
3 Complex numbers 3.1 Introduction . . . . . . . . . . . . . . . . . . . 3.2 Real and imaginary parts of z = x + yi . . . . 3.3 Geometric representation: the complex plane . 3.4 The complex conjugate and modulus . . . . . 3.5 Adding and multiplying . . . . . . . . . . . . 3.6 Real and imaginary parts . . . . . . . . . . . . 3.7 Solving equations involving complex numbers 3.8 Conjugates . . . . . . . . . . . . . . . . . . . . 3.9 Modulus . . . . . . . . . . . . . . . . . . . . . 3.10 Division . . . . . . . . . . . . . . . . . . . . . 3.11 The Mandelbrot set . . . . . . . . . . . . . . .
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62 62 63 63 63 64 65 66 68 68 70 72
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3.12 3.13 3.14 3.15 3.16
Argument . . . . . . . . . . . . . . . . . . Polar forms (modulus-argument forms) . . Multiplication and division of polar forms De Moivre’s theorem . . . . . . . . . . . . Roots of a complex number . . . . . . . .
4 Polynomials 4.1 Introduction . . . . . . . . . . . . . 4.2 Sketching polynomials . . . . . . . 4.3 Dividing one polynomial by another 4.4 Synthetic division . . . . . . . . . . 4.5 The remainder theorem . . . . . . . 4.6 Factors and roots . . . . . . . . . . 4.7 Real polynomials . . . . . . . . . . 4.8 Rational roots of a real polynomial 4.9 Repeated roots . . . . . . . . . . . 4.10 Final examples . . . . . . . . . . .
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73 75 77 79 80
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A Some common algebraic mistakes
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B Basic trigonometry
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0
Course Information for MATH160/101/102
MATH 160 has two parts: Algebra (MATH101) and Calculus (MATH102). Algebra and Calculus are used to model how things change. For example, to model the movement of an object through space, we create an algebraic structure in which to specify where the object is, and then we study how its position changes with time using calculus.
Lecturers 1st semester: 2nd semester: Calculus 1st semester: 2nd semester:
Algebra
Dr Florian Beyer, room 218,
[email protected] Dr J¨org Hennig, room 215,
[email protected] Prof Astrid an Huef, room 232A,
[email protected] TBA
Your Final Mark Your final mark F is made up of a test mark T (see the section “Skills Tests” below), an assignment mark A (see the section “Homework Assignments” below), and an exam mark E from the final exam. The marks T , A and E are all out of 100. Then F will depend on the formula 2E E A T , + . F = + max 3 3 2 6 For example, if A = 80, T = 60 and E = 70, then 140 70 80 60 F = + max , + ' 20 + max(46.67, 48.33) = 20 + 48.33 = 68.33. 3 3 2 6 On the other hand, if A = 70, T = 60 and E = 80, then 160 80 70 60 F = + max , + ' 20 + max(53.33, 51.67) = 20 + 53.33 = 73.33. 3 3 2 6 Note that 31 of your final mark is from the tests, up to 12 of your final mark can be from the tests and the assignments, and we use your assignment mark only if this helps you.
Terms Requirement “Terms” means the requirements a student must complete to be permitted to sit the final exam. If you fail Terms, then you fail this paper. To satisfy Terms for this paper you will need to: • “pass” (i.e. score at least 5 out of 10) on Skills Tests 1, 2, 3 and 4 for each of Algebra and Calculus. – MATH101 students take only the Algebra tests and MATH102 students take only the Calculus tests. • achieve an average mark of at least 40% on the weekly homework exercises. It is your responsibility to ensure that you meet these requirements.
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The MATH160/101/102 Webpages These webpages provide general and specific information about MATH160/101/102, and provide access to a Resources area, to the online homework assignments, and to your test and assignment marks. The instructions below are for MATH160; for 101/102 just substitute 101/102 for 160. (1) Go to the maths home page: www.maths.otago.ac.nz (2) Type math160 into the search field to get to the general MATH160 page. (3) Click on the Resources box. (4) You will see a list of downloadable files relevant to MATH160. (5) To access the online homework assignments and to see up-to-date details of your assessment and test marks, sign-in on the right side with your University username and a password. The password is initially your student ID number. (6) Change the password after you log in for the first time. (7) Click on “Continue to my personal resources page”. (8) This is where you will see available assignments and your marks. A shortcut to the above is to • Go to the math homepage www.maths.otago.ac.nz • Click “MATH160 resources” under “Quick links”. • Follow the instructions above from (5).
Homework Assignments There are 10 marked homework assignments, submitted and marked through an online system. To gain Terms you must achieve at least 40% of the available marks. Nevertheless, the homework assignments only contribute to your final mark if this helps you. The idea is that you can use the homework assignments to learn without too much penalty for getting things wrong as you learn. You can get lots of help for the homework assignments in the tutorials. • Instructions for accessing the MATH160/101/102 online assignment system are in the section “The MATH160 Webpage” above. You can access the system from any computer (in particular your computer at home) in contrast to the Skill Tests, which can only be done on the computers in room 124 in the Science III building at particular times. • Assignment 1 will be available on the Saturday in Week 1 from 1pm. You must enter all your answers by the Friday in Week 2 at 12pm (noon); at that time the system marks whatever has been entered. Your results will be available soon after this. In total, over the whole semester, there will be 10 marked assignments, each of which will be available from Saturday at 1pm to Friday 12pm (noon) of the following week. • Late submissions cannot be accepted. • In Week 1, a special assignment “Assignment 0” will be available. This assignment is not marked and hence does not contribute to your final mark. Nevertheless, it iv
is crucial that you attempt it by 12 pm (noon) on the Friday of Week 1 because it is your opportunity to practise entering the answers. You will need to know how to do this in all of the following 10 marked assignments later. • Before you attempt to answer any of the questions of Assignment 0, have a look at the instructions at the top of the page and also click on “Help with entering numbers and expressions”. This is where the syntax used by the system is explained and where you see lots of important examples of how to enter answers. The most important thing to remember when you enter answers is: – Give exact answers wherever possible, for example, 3.5, 5/11, 12+2/3, ln(2), sqrt(3)/2. If you must give an approximate answer, ensure it is accurate to at least 8 significant figures (unless the question says otherwise). Do not include units in your answers. • Is it important that you understand the steps involved in solving each assignment question since you will be required to show detailed working in the final exam! The exam consists of true/false and multiple-choice questions, as well as some long-answer questions where you have to show your working. So it is a good idea to work the assignment questions in a work book which you can consult again later.
Tutorials We run several open tutorial sessions each week, all in Room 124 of Science III. You are free to attend as often and for as long as you wish. The tutorials are there mainly to help you overcome any difficulties with the homework assignments, but you may also ask questions about material in the Skills Tests and the lectures. • Tutorials start in week 2. The tutorial times are: – 1st semester: Wednesday and Thursday 11:00 – 3:00; – 2nd semester: Wednesday and Thursday 10:00 – 12:00 and 1:00 – 3:00. It is important that you attempt the assignment before the tutorial, so that you already know what you need to ask about. Having looked at an assignment in advance, you will be able to ask the tutors for help with specific points. The tutorials are good places to form a study group. We encourage you to learn from each other.
The Skills Tests There are 5 Skills Tests in both Algebra and Calculus (MATH101 students only do the Algebra tests and MATH 102 students only do the Calculus tests). We give you a very good idea of what is on the tests via descriptions and/or sample questions available on the Resources page. The questions are not tricky or difficult, but are designed to make sure you have mastered the essential skills of the course. Your total score from all the tests will make up 31 of your final mark for the paper. To achieve Terms you must pass the first 4 tests in both Algebra and Calculus by the end of Week 12.
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• Each test must be done in room 124, Science III on one of the computers there. • For each test, there is an initial prescribed two- or three-week period in which to pass and gain full credit. During the prescribed period, you have up to four attempts, and your highest mark counts. For example, if you pass a test straight away with 7/10, then you have three more attempts to improve your mark; we count your best mark. Obviously, you should be aiming to maximise your test marks. • If you do not pass a test in the prescribed period, then you can continue to try to achieve the pass mark of 5 in order to get Terms, but the score that is recorded for the purposes of calculating your overall test mark will be limited. In the week immediately following the prescribed period this score will be 3; in subsequent weeks it will be 0. • Sessions times are Wed 4 pm – 7 pm, Thu 3 pm – 6 pm, Fri 9 am – 12 pm, 2 pm – 5 pm during the following weeks: Test Test Test Test Test
1 2 3 4 5
Semester 1 Weeks 3, 4, 5 (14 Mar - 8 April) Weeks 6, 7 (11 April - 22 Apr) Weeks 8, 9 (25 Apr - 6 May) Weeks 10, 11 (9 May - 20 May) Weeks 12, 13 (23 May - 3 June)
Semester 2 Weeks 3, 4 (25 Jul - 5 Aug) Weeks 5, 6 (8 Aug - 19 Aug) Weeks 7, 8, 9 (22 Aug - 16 Sep) Weeks 10, 11 (19 Sep - 30 Sep) Weeks 12, 13 (3 Oct - 14 Oct)
• The Algebra and Calculus tests are independent, and may be taken in different sessions. You may take both an Algebra and a Calculus test in the same session (but no more than that). • Test n + 1 cannot be attempted until Test n has been passed. • We will remind you of test times and availability in lectures, but the obligation to ensure that tests are not missed rests with you. • Email reminders will be sent to you also. So it is essential that you read your university email regularly. You may wish to set up automatic forwarding of messages to another email address. • If problems arise and you cannot take a test, then it is essential that you discuss these with one of the lecturers as soon as possible. What happens at the test? • Each test is done on a computer using specially-written software. Each has 10 questions. Tests are randomized and so each one is essentially unique. You have 12 minutes to complete a test. • Tests are held in the computer lab in Room 124 on the first floor of Science III. If you arrive early, please queue outside the room, as there could be another class being held. • Bring your student ID card and show it to the lab supervisor who will give you a 4-digit password. • Choose a computer and enter your ID number and the password you were given by the supervisor. vi
• You will now be looking at a summary of your test profile: which tests are available, which you have “passed”, your best mark, and which you can take now. Select the test you wish to take. • Read and follow the instructions. Select “Start The Test” when you are ready to begin and your time will start. • You must not use any reference material: the lab supervisor will check. Blank sheets of paper will be provided. You must not use a calculator. You must work on your own, with no talking. • You can review your answers, move back and forward from question to question, and if necessary make corrections at any time. At the end of the test, when you are satisfied with your answers, select “Exit”. You will be shown which questions you got right and your total score, and you can review the questions that you got wrong (for a maximum of 2 minutes). • If you want to take another available test then repeat the above. Otherwise select “Log out” and leave the computer at the sign-in screen for the next student.
Absences and Impaired Performance If an illness, accident or absence is serious enough to prevent you from submitting a homework assignment or from taking one of the tests, you must explain the circumstances to your lecturer. You may be asked to provide a medical certificate or other appropriate evidence. Provided you have a genuine reason, we can adjust your score so that you are not penalized. If for any reason you are absent from lectures, it is your responsibility to obtain the notes you have missed. If you don’t know anyone else in the class, go to a tutorial and seek help there. A tutor can help you to borrow the notes from another student for a few minutes so that you can make a copy.
Workload Since this is an 18-point, one-semester paper, you should expect to spend on average about 12 hours per week on it. This time might be made up from: • 5 hours attending lectures; • 5 hours on the exercises (including the tutorial) and the tests; • 2 hours reading your lecture notes or a textbook.
Mathercize The on-line system Mathercize allows you to test your background skills (e.g. material covered in High School) as well as to master and review material from MATH160. Familiarity with Mathercize may also help you with the Basic Skills Tests which use similar software. It is available on-line at http://mathercize.otago.ac.nz using a recent browser. You will need to supply your ID number and the Mathercize password “curve”.
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Textbooks and Outline Notes The MATH160 Algebra Outline Notes are available from the Print Shop and from the Resources page if you wish to print your own. Due to time constraints, the MATH160 Calculus Outline Notes are only available from the Resources page - please print them one-sided. For Calculus, the required text is either • MATH160 Calculus by James Stewart (truncated edition), available from the University Book Shop; or • Calculus by James Stewart (metric version, edition 8e), available from the University Book Shop. If you are planning on taking MATH170, you should consider getting the full Calculus rather than the truncated version. Reading out of Stewart will be assigned, and a list of extra suggested exercises from Stewart will be given. Those owning an older version of Stewart (or some other Calculus text book) can find the relevant sections easily. We will make available a photocopy of the extra problems. No additional textbook is required for Algebra. are:
Examples of suitable references
• An Introduction to Algebra and Vector Geometry by G.M. Kelly • Algebra, Geometry and Trigonometry in Science, Engineering and Mathematics by M.V. Sweet • Elementary Vector Algebra by A.M. Macbeath The above books are on reserve in the Science Library. They contain more advanced material than this paper and will include many more examples, both worked and otherwise, than we can cover in class.
Plagiarism Students should make sure that all submitted work is their own. From the University website: “Plagiarism is one form of dishonest practice. Plagiarism is defined as copying or paraphrasing another person’s work and presenting it as one’s own - whether intentionally, or through failure to take proper care. Being party to someone else’s plagiarism (by allowing them to copy your work or by otherwise helping them plagiarise work for an assessment) is also dishonest practice.” Any student found to be responsible for plagiarism in any piece of work submitted for assessment will be subject to the University’s dishonest practice regulations which may result in various penalties, including forfeiture of marks for the piece of work submitted, a zero grade for the paper or in extreme cases exclusion from the University. • We are happy for students to work together on the assignments, but you must submit your own work. • Strict observance of exam conditions (no talking, no reference materials, no calculator) is expected during the Skills Tests.
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1
Vectors
1.1
Introduction
The algebra of vectors arose out of the attempt to apply symbolic methods to geometry and mechanics. Many physical quantities can be described using a single number. Examples are temperature, time, mass, volume, pressure, electric charge. Such quantities are called scalars: they have magnitude but no direction. They obey the rules of algebra that we follow when manipulating real numbers. If the air temperature rises 10◦ between 9 a.m. and 1 p.m., and 5◦ between 1 p.m. and 6 p.m., then it has risen 15◦ between 9 a.m. and 6 p.m. — we just add the two numbers 10 and 5. However, there are other physical quantities which are not completely defined by their magnitudes. For example, if two forces F1 and F2 act on a body, we cannot conclude that the total effect is a force of magnitude F1 + F2 ; in fact, the two forces may be working against each other and hence cancel out. Clearly the directions of the forces are important.
F1
F2
If we move a body from point A to point B and then from point B to point C, then it has been displaced by distances AB and BC. But clearly its displacement AC from the original position is not AB +BC (unless A, B and C are in a straight line).
C
A B
Forces and directed displacements obey a different law of algebra from scalars. If we want F = F1 + F2 , and AC = AB + BC
(1.1)
to be correct, then we need to take into account both the magnitudes and directions of these quantities — we need vectors! Vectors are combinations of magnitude and direction, and obey an addition rule similar to that in (1.1). We can represent vectors by arrows:
v
This is the vector v
– the length gives the magnitude – the orientation gives the direction – the arrow-head gives the sense 1
a b c
Definition A non-zero vector v has: • magnitude, denoted |v|, which is a positive number; • direction in space: its orientation in space; • sense: specifies which end to draw the arrow-head. The zero vector 0 has magnitude zero (its direction and sense are undefined). We can now use a vector to represent a force (say): the above three properties describe the force completely. Equality We say that two vectors, a and b, are equal (and we write a = b) if they have the same magnitude, direction and sense.
Note that vectors do not depend on their position or placement in space. Notation We may also use points in space to denote the end-points of a vector. Thus the vector v linking point P to point Q is denoted PQ. Note that PQ and QP are different vectors.
P
v
Q
2
Sum of 2 vectors In physical situations we want to know the result of adding two vectors. For example, what is the resultant force when two forces act on a body? (Mathematically we can define the addition of vectors in many ways, but only one way is “natural” in that it is of general use in physical situations.) Definition. If a and b are two vectors represented by line segments PQ and QR, then a + b is the vector represented by PR.
R
P
a+b b
a
i.e. PQ + QR = PR
Q
Often called the triangle law of vector addition. Note that if points P and R are the same then we have: PQ + QP = PP = 0
(the zero vector)
QP is the negative of PQ. That is, QP = −PQ. Definition. The vector −a, the negative of a, has the same magnitude and direction as a, but opposite sense.
a a
Subtraction As with real numbers, we subtract a vector by adding the negative vector. That is, a − b means a + (−b). This process lets us solve equations:
Example 1.1. Construct the vector a − b.
a b 3
1.2
2- and 3-dimensional vectors
Most vectors representing physical quantities can be pictured in 2D or 3D space. We can specify vectors using components: x and y components in 2D space, x, y and z components in 3D space.
or
So (2, 1) is a 2-dimensional vector. Starting at any point P in the x-y plane, we move: 2 units in the x-direction, 1 unit in the y-direction to reach the point Q. Then the vector or, equivalently, or
(2, 1)
y Q 1
T P
S
2
R
= PQ = ST = OR
O
x
Note that the point R has coordinates (2, 1). (1, 2, 3) is a 3-dimensional vector. Starting at any point P we move: 1 unit in the x-direction, 2 units in the y-direction, 3 units in the z-direction, to reach the point Q. Then the vector or, equivalently,
(1, 2, 3)
Q
1 P
= PQ = OV
where point V has coordinates (1, 2, 3).
2
x
z V
3
O y
Addition In agreement with the triangle law of addition, vectors given in component form are added component by component. Thus: (1, 2, 3) + (0, −2, 4) = (1, 0, 7) We illustrate why this is true using 2D vectors:
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Position vector (p.v.) In the 2D example above, we would say that r = (2, 1) is the position vector of the point R. That is, r = OR. Similarly, in the 3D example, we would say that v = (1, 2, 3) is the position vector of the point V . Definition. The position vector p of a point P is that vector which, when placed with its foot at the origin O, has its head at P .
P Note. Applying the triangle law to the triangle OP Q, we have OP + PQ = OQ i.e. p + PQ = q
p O
q
Q
so: PQ = q − p, where p and q are the p.v.s of points P and Q. Warning: Be careful to distinguish between position vectors and other vectors. For example, in the triangle P QR, the vector a forms one side of the triangle — it is NOT the position vector of Q. In terms of Q position vectors we would have:
a
a = PQ = q − p while the vector a will be the position vector of some other point A.
1.3
P
b
R
Parallelograms
A parallelogram has opposite sides equal in length and direction. Thus in terms of vectors:
Q P
P QRS is a parallelogram if and only if PQ = SR.
R
An equivalent requirement is PS = QR. (Note the convention of labelling the vertices in order around the parallelogram.)
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1.4
Scalar multiplication
The sum a + a + a clearly forms a new vector with the same direction and sense as a but with 3 times the magnitude. It is natural to write it as 3a. Similarly, if k is a scalar, then ka is a vector with:
a
a
a
• the same direction as a, • magnitude |k| times that of a, • same sense as a if k > 0, opposite sense if k < 0. Component form When multiplying a vector by a scalar, we just multiply all components by that scalar. Thus:
3(1, 2, −3) = (3, 6, −9),
−2(5, −1, 4) = (−10, 2, −8)
Example 1.2. Given the vectors below, find (draw) a + 3b and 2a − 2b.
a b
Example 1.3. If 3x + a + 5b = x − a + b, then find x.
Example 1.4. If a = (1, 2, 0) and b = (1, −3, 6), then: 5a + 2b = 5b − 2a =
6
Example 1.5. Label these vectors with x, y, z in such a way that x = y + z.
Example 1.6. Label these vectors with p, s, t in such a way that p = s − t.
Example 1.7. Points A, B, C, D have position vectors a = (1, 4, 0), b = (2, 6, −3), c = (5, 1, −4), d = (4, −1, −1). (i) Show that ABCD is a parallelogram.
(ii) If BCAE is a parallelogram, then find the position vector of E.
P
Example 1.8. Given these three equilateral triangles, and the two vectors u and v as marked, find the following vectors in terms of u and v.
Q v
T
u S
R
(i) TP (ii) QR (iii) PR (iv) QT 7
1.5
The magnitude
The magnitude (length), |v|, of the vector v = (x, y) is given by Pythagoras’ theorem: 2
2
|v| = x + y
v x
so that: |v| =
y
2
p x2 + y 2
A similar formula holds for 3D vectors. If v = (x, y, z) = OP, then P has coordinates (x, y, z). Let Q be the point (x, y, 0). OQ2
Then
P (x, y, z)
= x2 + y 2
and in triangle OP Q (which has a right angle at Q)
So:
z
OP 2
= OQ2 + z 2 .
OP 2
= x2 + y 2 + z 2
O x y
and hence |v| = |OP| =
p
Q (x, y, 0)
x2 + y 2 + z 2
Example 1.9. Find |a|, |b| and |c| where a = (−1, 3, 4), b = (7, −4, 4), c =
b . 9
Example 1.10. Find the vector v with the same direction and sense as u = (2, 6, −3) but with magnitude 3.
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Example 1.11. The triangle ABC has vertices with p.v.s a = (1, −1, 1), b = (2, −1, 2), [ is a right angle. c = (2, 1, 0). Show that angle BAC
1.6
Unit vectors
A vector with magnitude = 1 is called a unit vector. 1 1 1 2 2 √ , 0, − √ , For example, (1, 0, 0), ,− , are all unit vectors. 3 3 3 2 2 b b In Example 1.9 we saw that |b| = 9 and = 1, so that is a unit vector. If we take 9 9 any non-zero vector a and divide it by its magnitude then we obtain a unit vector. We have: If a is a non-zero vector then: b a= (−
a is the unit vector with the same direction and sense as a. |a|
a is the unit vector with the same direction but opposite sense to a.) |a|
b which has the same direction and sense as Example 1.12. Find the unit vector v v = (1, 2, −3).
Example 1.13. Find the unit vector which has the same direction and opposite sense to v = (−4, 0, 3).
9
1.7
Parallel vectors
Vectors which have the same direction (irrespective of sense) are called parallel.
e.g.
ka is always parallel to a (for any non-zero k and a). Result. If a and b are parallel, then there is a scalar k 6= 0 such that a = kb. (That is, if two vectors are parallel, then one must be a non-zero scalar multiple of the other.)
a e.g.
2a
1 2a
Warning: The equation a = kb may or may not have a solution. a We cannot write k = . DIVISION BY VECTORS IS NOT DEFINED. b Example 1.14. Which of a = (1, −2, 4), b = (3, 6, 12), c = (−1.5, 3, −6) are parallel?
Example 1.15. Find the value of c so that (c, 8, −10) is parallel to (−2, −c, c + 1).
Example 1.16. Find the value of j so that (2, j, 3) is parallel to (2j, 3, j − 1).
10
Test for collinearity If three given points A, B, C are collinear — i.e. lie in a straight line — then AB and BC (or AC) will be parallel. So we have this test: Points A, B, C are collinear if and only if AB and BC are parallel. Example 1.17. Determine whether the points (5, 9, −2), (0, 5, 7) and (50, 45, −83) are collinear.
1.8
Applications to geometry
Section formula Suppose X lies on the line between points P and Q so that P X : XQ = α : β. Then the vectors PX and XQ are parallel and have the same sense.
X P
β
Q
α
In fact:
giving the section formula: x=
αq + βp α+β
Midpoint The midpoint M of the line joining P to Q has position vector: m=
p+q 2
11
Example 1.18. Let P and Q have position vectors (10, 50, −9) and (−11, 8, 12). (i) Find the position vector of the point X which divides P Q so that P X : XQ = 1 : 2.
(ii) Find the point Y which divides XQ so that XY : Y Q = 3 : 4.
(iii) Find the midpoint M of XY .
Centroid of a triangle Example 1.19. Consider the triangle ABC. Let points A, B, C have p.v.s a, b, c, and let M be the midpoint of the side BC. Find the p.v. of G, the point on AM such that AG : GM = 2 : 1.
12
1.9
The scalar product
If a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ), then we know that
and
|a|2
=
|b|2
=
So |a − b|2 = |(a1 − b1 , a2 − b2 , a3 − b3 )|2 = (a1 − b1 )2 + (a2 − b2 )2 + (a3 − b3 )2 = a1 2 + b1 2 − 2a1 b1 + a2 2 + b2 2 − 2a2 b2 + a3 2 + b3 2 − 2a3 b3 = =
(1.2)
The term a1 b1 + a2 b2 + a3 b3 crops up often when we deal with lengths of vectors in 3D space, and so we give it a special name and notation: Definition. If a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ), then a · b = a1 b1 + a2 b2 + a3 b3 is the scalar product or dot product of a and b. Why “scalar product”? Because a · b is a scalar, not a vector. (There is also a vector product, denoted a × b. This is not covered in Math160.)
Example 1.20. If a = (1, 5, −2), b = (0, 2, 1) and c = (4, −2, −3), then a·b=
a·c=
a·a=
13
Properties The scalar product obeys the following laws: Commutative:
a·b=b·a
Distributive:
a · (b + c) = a · b + a · c
and a · a = |a|2 What about the associative law?
We can now use the above laws to recalculate |a − b|2 :
(1.3) Geometric interpretation: Consider this triangle: where θ is the angle between the vectors a and b. The cosine rule of trigonometry tells us that
a
a
b
θ b
|a − b|2 = |a|2 + |b|2 − 2|a||b| cos θ Comparing this with (1.3) gives us that a · b = |a||b| cos θ.
(1.4)
We thus have two expressions for the scalar product: a · b = a1 b 1 + a2 b 2 + a3 b 3 a · b = |a||b| cos θ θ is the (smallest) angle of rotation needed to bring the direction and sense of a into line with those of b.
14
If a and b are perpendicular (or orthogonal) then θ = 90◦ , so cos θ = 0 and a · b = 0. If a · b = 0 then what can we conclude? From a · b = |a||b| cos θ we can say that either a = 0 or b = 0 or θ = 90◦ . So: If a and b are non-zero vectors then a·b=0
⇔
a and b are perpendicular.
Example 1.20 revisited: The vectors a and c are perpendicular.
Example 1.21. The triangle ABC has vertices with p.v.s a = (1, −1, 1), b = (2, −1, 2), [ is a right angle. c = (2, 1, 0). Show that angle BAC (We showed this in Example 1.11 by using Pythagoras’ theorem.)
Example 1.22. Find u · v given that u is a unit vector, v has a magnitude of 6, and u and v have the same direction but opposite sense.
15
Angles The scalar product gives a means for finding the angle between two vectors. From (1.4) we have that cos θ =
a·b |a||b|
If cos θ > 0 then θ is an acute angle (between 0 and 90◦ ). If cos θ < 0 then θ is an obtuse angle (between 90◦ and 180◦ ). If we definitely want the acute angle between two directions then we must take the absolute value of the scalar product; that is: |a · b| For acute θ, cos θ = |a||b|
Example 1.23. Find the angle between the vectors u = (2, 2, −1) and v = (1, 4, 1).
Example 1.24. Find the acute angle between the directions d = (−2, 1, 1) and e = (1, −1, 0).
◦ Example √ 1.25. Find a vector u of length 5 that makes an angle of 30 with the vector a = (3, 2, 1) and is perpendicular to b = (0, 1, 0).
16
1.10
The parametric equations of a line
We consider the straight line which passes through the point P (with p.v. p) and has direction d.
d
P
X
p O
x
If X is a general point on the line, then in the triangle OP X we see that OX = OP + PX But OX is the position vector of X, i.e. x, OP is the position vector of P , i.e. p, PX is parallel to the direction d, so PX = td, for some scalar t. Hence: x = p + td. The scalar t is called a parameter of the line, and this equation is called a parametric equation of the line through P with direction d. The parameter t can take on any real value. As it varies, the point X moves along the line. Some positions and corresponding values of t are shown here.
Example 1.26. Find a parametric equation of the line through the point (2, 0, 7) with direction (9, −1, −2).
If the position vector x of the general point X on the line has components (x1 , x2 , x3 ) and similarly p = (p1 , p2 , p3 ), d = (d1 , d2 , d3 ), then we can write the parametric equation in the component form: (x1 , x2 , x3 ) = (p1 , p2 , p3 ) + t(d1 , d2 , d3 ) or x1 = p1 + td1 x2 = p2 + td2 x3 = p3 + td3
(1.5)
17
Example 1.27. Let a = (1, 0, 1) and c = (0, 1, 2) be the p.v.s of points A and C. (i) Find a parametric equation of side AB of the parallelogram OABC.
(ii) Find the point X on the side AB of the parallelogram OABC so that OX is perpendicular to XC.
The same line can have many different parametric equations — in fact infinitely many. After all we can choose any point P on the line and there are infinitely many of those. The direction vector d can also be chosen from any vector parallel to the required direction. To show that two parametric equations represent the same line, we need to confirm that, (i) the two directions are parallel; (ii) there is (at least) one point whose position vector satisfies both equations. Example 1.28. Show that the parametric equations x = (1, 3, 0) + t(3, −1, −2) and x = (7, 1, −4) + t(−3, 1, 2) represent the same line.
18
Example 1.29. Find a parametric equation of the line through the two points P , Q with position vectors p = (1, 2, 5), q = (0, −6, 7).
The parameter as a measurer of distance Consider an equation of the line through the two points P and Q. As we saw in Example 1.29, a parametric equation is x = p + td where d = PQ = q − p, or x = p+t(q − p). Now when t = 0, x = p. When t = 1, x = So the part of the line between P and Q is given by 0 ≤ t ≤ 1. The parameter t here expresses the distance from P to X relative to the distance from P to Q. By scaling the direction vector so that it is a unit vector, we can make the parameter t give us an absolute measure ofdistance from P along the line. So instead of using PQ d i.e. . PQ as the direction, we use PQ, |PQ| Now when t = 0, x = p. And when t = |PQ|,
Example 1.30. Find the point at a distance 2 from the point P (1, 3, 0) along the line from P to Q (3, 1, 1).
Example 1.31. Find the point at a distance 3 from the point Z (1, 1, 4) in the direction (4, 0, −3).
19
Example 1.32. Find two points on the line through P (1, 3, 0) and Q (8, −1, 4) that are a distance 5 from P .
1.11
Projections
a
Suppose we have a vector a and a line with b (a unit vector). direction d Drop a perpendicular from each end of the vector a onto the line — let these meet the line at points P and Q.
d
Q P
b P Q is called the projection of a onto the direction d. What is the length of P Q? Draw an equal vector a = PR. Then in triangle P QR we have P Q = But also
b= a·d
So we have: b is a · d. b The projection of a onto direction d b must be a unit vector! Note that the direction d b is the signed length PQ. If we want just the usual length we take the The value of a · d b absolute value of the projection: |a · d|. Example 1.33. Find the projection of (5, 1, 2) in the direction of the vector (0, 3, 4).
Example 1.34. Find the projection of (1, 1, 3) onto the line x = (0, 2, 1) + t(−1, 2 − 2).
20
If we want the vector PQ (called the projection vector) then we multiply the projecb tion by the unit direction vector d: b is (a · d) b d. b The projection vector of a onto direction d Example 1.35. Find the projection vector of (1, 5, 10) in the direction (6, 2, −3).
1.12
Distance from a point to a line
We find the perpendicular distance from the point A to the line x = p + td.
d P ? A
r Distance =
2 b |a − p|2 − (a − p) · d
Example 1.36. Find the distance from the point A = (3, −2, 1) to the line through P = (4, 1, 3) parallel to d = (1, 1, 1).
Example 1.37. Find the perpendicular height of the triangle ABC.
A (2, 1, 2)
B (0, 3, 1)
C ( 2, 1, 0)
21
Example 1.38. Find the distance between the parallel lines through points (1, 0, 4) and (2, −1, 0) with direction (0, 1, 2).
1.13
The normal equation of a plane
First, consider what we mean by a plane: • A plane is a 2-dimensional surface in 3-dimensional space. • There is a unique direction perpendicular to the plane — a vector with this direction is called a normal vector of the plane.
n For example, if P is a fixed point on the plane, and X is another point on the plane, then all vectors PX are perpendicular to the normal vector n.
P
X2
X1
Note: It is only the direction of the normal vector which is important. There are many normal vectors that we could use, but they are all parallel. Since PX is perpendicular to n, we have that PX · n = 0. That is:
The equation of a plane with normal n is x · n = k. k can be found if we know a point on the plane. This equation is sometimes called the normal equation of the plane. 22
If n = (n1 , n2 , n3 ), then the plane has equation: n1 x1 + n2 x2 + n3 x3 = k or n1 x + n2 y + n3 z = k (depending on whether we are using components (x1 , x2 , x3 ) or (x, y, z)). Note that the components of the normal vector are just the coefficients in the equation. So these are all planes:
x + y − 2z = 3,
They have normal vectors n = (
2x − 5z = 100, ), n = (
y = 9. ), n = (
).
A point on the plane x + y − 2z = 3 is any point whose coordinates fit the equation. For example:
Example 1.39. Find a normal and a few points on the plane 3x − y + 4z = 9.
Example 1.40. (i) Find the equation of the plane through the point A (3, −2, 1) with normal (1, 2, 3).
(ii) Choose any other point B on the plane and confirm that AB is perpendicular to the normal.
23
Example 1.41. Find the plane through the origin which is perpendicular to the line x = (1 + 2t, 5t, −7 + 3t).
Example 1.42. Find the equation of the plane through the three points A (0, 1, −1), B (1, 1, 0), C (1, 2, 0).
1.14
Angle between two planes
The angle between two planes is taken to be the acute angle between their normal vectors. In fact it is the angle between two lines, one in each plane, and both perpendicular to the line of intersection of the planes.
n1
n2
plane 1
θ Line of intersection
θ plane 2
24
We can use n1 · n2 = |n1 ||n2 | cos θ to find the angle. Since we want the acute angle we take the absolute value of the scalar product, so really we have: cos θ =
|n1 · n2 | |n1 ||n2 |
Example 1.43. Find the angle between the planes 2x+2y +z = 0 and −x−2y +2z = 3.
1.15
Intersection of a line and a plane
To find where a line and a plane intersect, we must find the point where the equations of the line and plane are both satisfied. It is easiest to start with the parametric equation of the line. The method is as follows: • Start with a parametric equation of the line x = p + td. • Write out the components in terms of the parameter t. • Substitute these into the equation of the plane. • Solve for t. • Put this value of t back in the parametric equation of the line to find the point. Example 1.44. Find where the line through (1, 0, 4) with direction (2, 1, −1) meets the plane 3x − y − 2z = 9.
Example 1.45. Find where the line x = (1, 2, 0) + t(2, 0, 3) meets the plane −x + 5y + z = 12.
25
Example 1.46. Find where the line x = (1, −2, 4) + t(7, 10, 8) meets the plane z = 0.
1.16
Distance from a point to a plane
We find the perpendicular distance from the point A to the plane x · n = k. A
?
n
Distance =
|a · n − k| |n|
Example 1.47. Find the distances from the points (3, −2, 1) and (−4, 21, 3) to the plane 2x + y − 2z = 7.
26
Parallel planes Two planes are parallel if they have parallel normal vectors.
To find the distance between parallel planes just find the distance from a point on one plane to the other plane. Example 1.48. Find the distance between the planes −4x − 6y + 6z = 4 and
1.17
2x + 3y − 3z = 1.
Projections onto a plane
Projection of a point A Let us first find the foot Q of the perpendicular line from a given point A to a given plane x · n = k. Q is the projection of A onto the plane. All that is involved here is finding the intersection of the perpendicular line with the plane.
n Q
The line passes through A and has direction n. So we need to find the intersection of the line x = a + tn with the plane x · n = k — a problem we solved earlier. Example 1.49. Find the projection of the point (7, 2, 1) onto the plane 2x − 3y − z = 0.
27
Projection of a line segment A B
To find the projection P Q of the line segment AB onto a plane, we can use two methods.
Q
n P
1. Find the feet P and Q using the method above. Hence we can then find the length P Q. 2. If only the length of P Q is required (and not the positions of the points) we could argue as follows:
Example 1.50. Find the projected length of the line segment between points (1, −3, 4) and (2, 2, 7) onto the plane 3x + 4y + z = 8. Method 1:
Projection of A = (1, −3, 4): Line is x = (1, −3, 4) + t(3, 4, 1) = (1 + 3t, −3 + 4t, 4 + t). This hits plane where 3(1 + 3t) + 4(−3 + 4t) + (4 + t) = 8
Projection of B = (2, 2, 7):
Method 2:
28
1.18
The parametric equation of a plane
In some circumstances it is beneficial to describe a plane, not by its normal equation x · n = k, but by a parametric form similar to the parametric equation of a line. Consider the plane that passes through the point P (with position vector p) and that is parallel to the two vectors u and v. We can think of u and v actually lying in the plane as shown. We will assume that u and v are not parallel.
Then any point X in the plane has position vector
We have derived a parametric equation of the plane: x = p + su + tv Note that this has two parameters, s and t. Compare this to the parametric form of a line x = p + td which has one parameter. This fits in with the ideas that a line is one-dimensional and a plane is two-dimensional. It is usual to choose u and v as perpendicular unit vectors as this makes finding s and t simpler. If this is the case then u · v = 0, u · u = 1, v · v = 1, so x·u= and x·v =
Hence: If u and v are perpendicular unit vectors then s = (x − p) · u and t = (x − p) · v
29
Example 1.51. Find a parametric equation of the plane 2x + 3y − z = 5.
1 Example 1.52. For the plane 2x − y + z = 4 confirm that u = √ (0, 1, 1) and 2 1 v = √ (1, 1, −1) are perpendicular unit vectors in the plane. 3 Given that points P (1, −1, 1) and X (3, 3, 1) lie on the plane, express x in terms of p, u and v.
30
1.19
An application of vectors to calculus
The vectors we have been studying so far have been mostly fixed — for example, (1, 2, 3) or (0, −2, 4). A vector of the form (1, 2, 0) + t(−1, 3, 2) or equivalently (1 − t, 2 + 3t, 2t) varies with the parameter t and of course gives the position vector of a general (or moving) point on a straight line. If t represents time, then the expressions for the three components of such a vector are linear functions of time: 1 − t, 2 + 3t and 2t. In applied mathematics we often study an object moving in 2- or 3-dimensional space. Its position at any particular time t can be described using the position vectors r = (x, y) or r = (x, y, z) where x, y and z are functions of t.
Consider a particle at point A with position vector r(t). Then, a little time h later, suppose it is at point B with position vector r(t + h). So r(t + h) − r(t) = AB. We define the derivative of r(t) as:
As h → 0, B → A and AB becomes parallel to the tangent at A.
In fact v =
dr gives the velocity of the particle. dt Its direction is always tangential to the curve and is the direction of travel at time t. Its magnitude gives the speed.
d2 r Then a = 2 is the acceleration of the particle. dt dr We calculate by differentiating each component of r, i.e. dt dr dx dy dz = , , . dt dt dt dt
31
Example 1.53. Let the particle move along the curve given by r = (1 + t, 2 − 3t, 2t1.5 ) for t ≥ 0.
Since |v| = speed at time t, we can find the distance travelled (which equals the arclength of the path) by integrating. Z t2 |v| dt. That is, the distance travelled in the time interval [t1 , t2 ] is t1
Example 1.54. Let the path be given by r = (t2 , 2t, ln t) for t > 0.
Example 1.55. The path given by r = (cos t, sin t, t) is a helix.
32
1.20
An application of vectors to computer graphics
In 3D computer graphics we view objects projected onto the computer screen or what we will call the view-plane. We will assume that the eye is looking toward the origin — the view-plane is perpendicular to this line of sight. If N is the point where the line of sight meets the view-plane, then let n = ON. Then the view-plane has equation:
Points on some 3D object are projected onto the view-plane by one of two types of projection. Perpendicular projection Here we project points perpendicularly to the plane — we can use the usual method for projecting points onto a plane. Perspective projection This type of projection is usually preferred as it makes closer objects look larger than distant objects. Here we construct a line from a point on the object to the eye (which might be, for example, at the point 3n). We then take the intersection of that line with the view-plane. Example 1.56. Find the image of the point A (1, 3, −1) onto the view-plane at n = (3, 6, 3) using both methods. Assume that the eye is at 3n.
33
Screen coordinates Having projected an object onto the view-plane we need to switch to a new coordinate system in the plane itself in order to draw the image. Suppose u and v are perpendicular unit vectors lying in the plane (so they are also perpendicular to the normal vector n). They form what is called a basis for the plane. If x is the position vector of a point X on the plane then we can write x = n + su + tv Then
x·u=
So in screen coordinates (with the point N as origin) the point X is at (s, t). 1 Example 1.57. (continuing example 1.56) Here n = (3, 6, 3). Take u = √ (2, −1, 0), 5 1 v = √ (1, 2, −5). These three vectors are mutually perpendicular, and u, v are unit 30 vectors.
34
2 2.1
Linear equations and matrices Systems of linear equations
A linear equation in the n variables (unknowns) x1 , x2 , . . . , xn is one which can be expressed in the form a1 x 1 + a2 x 2 + · · · + an x n = b where a1 , a2 , . . . , an and b are constants. (The word linear comes from the fact that when n = 2, the equation a1 x1 + a2 x2 = b is that of a line.) If we only have a few variables, we will frequently avoid subscripts and call them x, y, z, . . . instead of x1 , x2 , x3 , . . . For example, the following are linear equations: x + 3y = 7,
x1 − 2x2 − 3x3 + x4 = 7,
x + 3y + 2z + w = 1.
A linear equation does not involve products or roots of variables, and each variable occurs only to the first power and not as the argument of a trig, log or exponential function. So the following are not linear: x + 3y 2 = 7,
x1 − x1 x2 +
√
x3 = 3,
y − sin(x) + ez = 2.
A finite number of linear equations which we solve simultaneously is called a system of linear equations. A combination of numbers s1 , s2 , . . . , sn is called a solution of the system if x1 = s1 , x2 = s2 , . . . , xn = sn is a solution of every equation in the system. For example, the system 4x1 − x2 + 3x3 = −1 3x1 + x2 + 9x3 = −4 has solution x1 = 1, x2 = 2, x3 = −1, since these values satisfy both equations, whereas x1 = 1, x2 = 8, x3 = 1 is not a solution since these values satisfy the first equation but not the second. A system may have more than one solution (e.g. x1 = 5, x2 = 11, x3 = −3 13 is also a solution of the system above); all possible solutions s1 , s2 , . . . , sn to a system is called the solution set.
35
Recall that given a pair of simultaneous equations in two variables the corresponding lines lie in a plane and may be: (i) parallel,
(ii) intersecting,
(iii) coincident.
That is, the pair of equations has either no solution, exactly one solution, or infinitely many solutions. Put another way, the solution set is either empty, has one element, or infinitely many elements. We will see later that the same applies for any system of linear equations.
2.2
Augmented matrices
Associated with a system of linear equations we can write down its augmented matrix. We will look more generally at matrices later, but for now we regard a matrix as a rectangular array of numbers, enclosed in a pair of brackets. The augmented matrix of a system of linear equations contains all the coefficients and right-hand sides of those equations, without the variables. It is just another way of displaying the important numbers from each equation. For example: The system
3x + 4y = 9 2x − 7y = 5
has the augmented matrix
The system
x + 4z = 7 2x − y − z = 4 3y − 3z = −5
has the augmented matrix
3 4 9 2 −7 5
Similarly, assuming the variables are x, y and z, the augmented matrix 1 1 0 2 −3 2 2 1 is just another way of writing the system: 2 2 7 −3
Note that each row of the matrix corresponds to one linear equation.
36
2.3
Operations on equations
To solve a system of equations using the standard method of eliminating variables we allow ourselves 3 natural operations: Type I: Multiply any equation by a nonzero constant. Type II: Add a multiple of one equation to another. Type III: Interchange two equations. Example 2.1. Using the operations above, solve the system S of simultaneous equations: 3x + 6y = 39 2x + 5y = 31
(2.1) (2.2)
Here we didn’t use a type III operation, but it is useful in larger systems. The idea used above was to replace the first system of equations S by another system S 0 , which is easier to solve. Note however that the solution to each system is the same. Geometrically we have:
The steps in the solution in Example 2.1 could have been performed on the rows of the augmented matrix rather than on the equations themselves. The operations of types I, II, III translate to the following “row operations” on an augmented matrix: Type I: Multiply a row by a nonzero constant. Type II: Add a multiple of one row to another. Type III: Interchange two rows. 37
Example 2.2. Apply row operations to the augmented matrix of the system S of Example 2.1 to reduce it to an equivalent system S 0 which is easier to solve.
2.4
Equivalent systems
Suppose S is a system of linear equations. A system S 0 of equations is equivalent to the system S, if S 0 can be obtained from S by a finite number of operations of type I, II or III (above). Result. If S and S 0 are equivalent systems of equations then S and S 0 have the same solution set. (Any solution of system S is a solution of S 0 and vice versa.) Justification It’s obvious that operations I and III don’t affect the solution set. Consider the system: 2x + 3y = 12 x + 4y = 8
(2.3) (2.4)
Scaling the equations (type I operations):
or swapping them (a type III operation)
doesn’t in any way change the two straight lines that correspond to these equations, and hence doesn’t change where they intersect (the solution). A type II operation like replacing (2.4) with (2.4) + 2 (2.3) does appear to change one equation, but if x and y are values that satisfy (2.3) and (2.4), then they must also satisfy the newly formed equation — these same values will also be a solution of the new system: (2.5) (2.6) It is also clear that we haven’t introduced a new solution as this process is fully reversible: we just replace equation (2.6) with (2.6) +(−2) (2.5). 38
Notation. We say augmented matrix B is equivalent to A, and write A ∼ B, if B is obtained from A by a sequence of row operations. (Of course B ∼ A also, since we can easily reverse the operations to get from B back to A.) In performing row operations it is important to explain how we get from one matrix to the next. At each step we do this by specifying at the end of each row that has changed in the new matrix, how it was obtained from the rows of the previous matrix. We use the greek letter ρ (rho) and denote by ρ1 , ρ2 , . . . , the first row, second row, . . . , of the previous matrix. Example 2.3. Repeat Example 2.2 showing equivalences and the ρ notation.
2.5
Gauss reduction
As in Example 2.3 above, we always aim to finish up with a simple matrix of the following type: Reduced row echelon form: A matrix is in reduced row echelon form (or is a reduced row echelon matrix ) if it satisfies the following properties: (i) all the zero rows (if any) are at the bottom of the matrix; (ii) the first nonzero entry in every nonzero row is a 1 (called the leading entry of its row); (iii) the leading entries move progressively to the right going down the matrix; (iv) the leading entry in a row is the only nonzero entry in that column. The word “echelon” means “step-like”. A matrix which satisfies the first 3 properties but not condition (iv) is said to be a row echelon matrix. For example,
1 0 0 2 1 0 2 0 3 0 1 3 0 A = 0 1 0 3 , B = 0 1 0 0 −2 , C = 0 0 0 1 0 0 1 −5 0 0 0 1 0 0 0 0 0 are all in reduced row echelon form.
39
However, 0 D= 0 0 1 E = 0 0 0 1 F = 0 0 1 0 G= 0 0 1 H= 0 0
these are not in reduced row echelon form: 1 0 3 0 0 0 0 1 5 0 1 0 2 1 0 0 0 1 1 1 0 0 3 0 0 0 1 0 0 0 3 1 1 3 0 1 0 0 2 −1 −2 3 0 1 −6 2 0 0 1 0
(The matrices G and H are in row echelon, but not reduced row echelon, form.) Our aim now in solving a system of equations is to reduce the augmented matrix of the original system to a reduced row echelon matrix, from which it is easy to read off the solution.
For example, from A above we easily get the solution: x = 2, y = 3, z = −5. We can do this reduction process because of this result: Result. A matrix is equivalent to a unique reduced row echelon matrix. That is, it doesn’t matter which row operations we do and in which order, the final reduced row echelon form will always be the same for that given matrix. The reduced row echelon matrix is produced by applying a sequence of row operations to the original augmented matrix. The step by step algorithm for doing this is called Gauss reduction (or Gaussian elimination). There are several different versions of this, but all versions produce the same reduced row echelon form. The version we will use follows this idea: Over-riding idea: we process the matrix column by column, from left to right, transforming each column into the required form of the reduced row echelon matrix.
40
Normally we need to process ALL but the last column of the augmented matrix. Occasionally, however, we are able to skip a column when entries on and below the diagonal are all zero already: we then just move on to the next column. However, for now we will assume that we are working with a system of n equations in n unknowns and that nothing unusual happens. For the kth column, we want: (i) 1 in the kth row. To get this, we may need to: • interchange the kth row with another row (a Type III operation) • scale the kth row (a Type I operation) (ii) 0s above and below that 1. We may need to: • add multiples of the kth row to the other rows (Type II operations)
So the 1st column will have a 1 in the 1st row and 0s under it:
The 2nd column will have 1 in the 2nd row and 0s above and below:
And so on.
0 2 4 6 The method is illustrated on the matrix: 3 −3 6 15 . 2 5 7 9 1st column (i.e. k = 1) We need to interchange two rows to bring a non-zero entry to the first row: inter-change rows 1 and 2 (say). Then scale the new 1st row by dividing through by 3:
Next, produce 0s below the 1. There is already a 0 in the 2nd row. We produce a 0 in the 3rd row by adding (−2) times the 1st row to the 3rd row:
41
2nd column (i.e. k = 2) The entry in the second row of this column is already non-zero, so no interchanges are needed. However we need to scale the second row:
Now produce 0s above and below the 1 by adding suitable multiples of the 2nd row to the other rows:
3rd column (i.e. k = 3) The entry in the third row of this column is already non-zero, so no interchanges are needed. However we need to scale the third row:
Now produce 0s above the 1 by adding suitable multiples of the 3rd row to the other rows:
The matrix is now in reduced row echelon form. The corresponding equations are:
which constitutes the solution. To check our working we should now substitute these values back into the original equations and make sure that ALL are satisfied. Summary. Procedure for solving a system of linear equations: 1. Obtain the augmented matrix of the system. 2. Do any obvious scaling of the equations to help later arithmetic. 3. Use Gauss reduction to obtain the equivalent reduced row echelon matrix. 4. Read off the solution from the row echelon matrix. 5. Check by substitution that the solution works. 42
Example 2.4. Use Gauss reduction to solve the system of equations: 2x − 3y = 1 x + 2y = 5
Example 2.5. Insert the row operations in the following solution of the system of equations: 2x + 4y + z = 5 x + 2y + 2z = −2 3x + y + z = 4
2 1 3 1 ∼ 0 0
4 1 5 2 2 −2 1 1 4 2 2 −2 0 −3 9 −5 −5 10
1 2 2 −2 ∼ 0 1 1 −2 0 0 −3 9
1 0 0 2 ∼ 0 1 1 −2 0 0 1 −3
1 ∼ 2 3 1 ∼ 0 0
2 2 −2 4 1 5 1 1 4 2 2 −2 −5 −5 10 0 −3 9
1 0 0 2 ∼ 0 1 1 −2 0 0 −3 9
1 0 0 2 ∼ 0 1 0 1 0 0 1 −3
43
Example 2.6. A person is ordered by his doctor to take 33 units of vitamin A, 50 units of vitamin B, and 63 units of vitamin C each day. The person can choose from three brands of vitamin pills. Brand X contains 3 units of A, 2 units of B, and 4 units of C, Brand Y contains 3 units of A, 4 units of B, and 5 units of C, Brand Z contains 3 units of A, 6 units of B, and 7 units of C. If the person meets his vitamin requirements exactly by taking x of brand X, y of brand Y and z of brand Z each day, write down equations which x, y and z satisfy, and solve these using Gauss reduction to bring the augmented matrix to reduced row echelon form.
1 1 −1 2 Example 2.7. Find the reduced row echelon form of the matrix 0 1 1 5. −1 1 3 8
44
2.6
Multiple solutions and inconsistent systems
If the matrix in Example 2.7 is the augmented matrix of a system of equations in variables x, y and z, then the reduced row echelon form corresponds to the equivalent system: x − 2z = −3 y+z =5 This system does not have a unique solution. In fact if we choose any value for z, and then let x = 2z − 3, y = 5 − z, we will have a solution. For example
Example 2.8. Find all solutions of the system of equations : x+y+z =4 x + 2y + 3z = 10
(2.7) (2.8)
The variables that correspond to the leading 1s are called leading variables (x and y in Example 2.8). Any nonleading variables there may be (such as z in Example 2.8) are called free variables, because they may be chosen in an arbitrary way. This means that: Any system with one (or more) free variables has infinitely many solutions. Geometrically (2.7) and (2.8) in Example 2.8 represent planes and the general solution found above is the line of intersection of the two planes. When the solution set is expressed in terms of one or more free variables, it is usually called the general solution of the system. For Example 2.8:
45
Example 2.9. Suppose the vitamin requirements and the numbers of units that each brand contained in Example 2.6 were different so that now x, y and z satisfy a system 1 1 2 10 of equations with augmented matrix: 0 2 2 6 . 1 4 5 19 Find all possible solutions for x, y and z.
1 ∼ 0 0 1 ∼ 0 0
1 2 10 2 2 6 3 3 9 0 1 7 1 1 3 0 0 0
1 1 2 10 ∼ 0 1 1 3 0 3 3 9
Note. Geometrically the 3 equations in Example 2.9 represent 3 planes that meet in a line. The general solution in fact is just the equation of this line. The next examples have two free variables: Example 2.10. Find the general 1 2 1 with augmented matrix 0 0 1 2 4 2
solution of the system of equations in x, y, z and w −1 3 −1 2. −2 6
46
Example 2.11. Writedown the general solution corresponding to the augmented ma 1 0 0 1 4 trix: 0 1 1 −2 7. 0 0 0 0 0
So far, any system we have considered has had either one solution or infinitely many solutions. We say such a system is consistent. We next examine a system which has no solution and is thus said to be inconsistent. Example 2.12. Show that the system of equations in x, y and z with augmented matrix given below is inconsistent. 1 2 1 1 2 5 2 3 3 5 3 4
If a row reduced augmented matrix contains a row 0 0 . . . 0 X where X 6= 0, then the corresponding system is inconsistent. Example 2.13. Describe geometrically the 3 planes represented by the 3 equations in Example 2.12.
47
Example 2.14. A group of children, teenagers and adults (15 in total) went to a fairground. On one ride, children were free, teenagers paid $1, adults $4, and the total paid was $24. On another ride, children paid $2, teenagers $3, adults $6, and the total paid was $57. Can we determine how many were in each age-group?
What if the $57 were changed to $54?
Summary. Inconsistent systems and multiple solutions: 1. Transform the augmented matrix to reduced row echelon form. 2. If there is a row 000 . . . 01 then the system is inconsistent. 3. If there are fewer non-zero rows than variables, then there are infinitely-many solutions. A row such as (1 0 1 0 4) translates to x = 4 − t where z = t, and t is a parameter in R (or in a subset of R determined from practical restrictions on the solution).
2.7
Matrices
Definition. A matrix is a rectangular array of numbers. The numbers in the array are called the entries or elements of the matrix. The size of a matrix is given by the number of rows and columns it contains. Example 2.15. Some examples of matrices are √ 4 6 2 1/2 3 0 −1 , 1 2 3 , 0 e2 2 , 2 3 0 0 1
π π/2
Their sizes are Note. The number of rows is always given first. 48
We use capital letters A, B, C, . . . for matrices and small letters for their entries. The matrix A has entries aij and we often write A = [aij ]. Subscripts are used to denote specific entries, again with row before column. aij is the entry in row i and column j of the matrix A So if A is the first matrix in Example 2.15, we have a22 = , a31 = 2 −3 8 Example 2.16. If A = , find the entries a12 , a22 and a23 . 10 88 −2
.
Example 2.17. Let A = [aij ] be a 2 × 3 matrix, where aij = i + j. (i) Write down the matrix A.
(ii) Find
3 P
a1j ,
j=1
2 P i=1
aii ,
2 P
ai3 .
i=1
More definitions A 1 × n matrix is called a row matrix or row vector.
An m × 1 matrix is called a column matrix or column vector.
An n × n matrix is called a square matrix.
A zero matrix is a matrix with all entries 0 and is denoted 0.
Two matrices A = [aij ] and B = [bij ] are equal if they have the same size (both m × n, say) and their corresponding entries are equal (aij = bij for all values of i and j). Example 2.18.
49
Transpose The transpose of an m × n matrix A, denoted AT , is the n × m matrix obtained from A by interchanging its rows and columns. That is, the first column of AT is the first row of A, the second column of AT is the second row of A, etc. Thus the (i, j) entry in AT is the (j, i) entry of A. 1 6 0 Example 2.19. Find the transpose of A = and B = −1 7 3 0 . −3 0 7
2.8
Matrix operations
Addition and subtraction Two matrices A and B can only be added when they both have the same size, say m×n. Then their sum A + B is the m × n matrix formed by adding the corresponding entries in A and B, and their difference A − B is the m × n matrix obtained by subtracting the corresponding entries in A and B. 2 −1 3 0 1 2 5 1 Example 2.20. Let A = , B = and C = . Find 0 5 2 2 −1 4 3 −2 (where defined) A + B, A − B, A + C, B − C.
Scalar multiplication If A is an m × n matrix and λ is a number (often called a scalar), then we denote by λA the m × n matrix in which each entry of A is multiplied by λ. Example 2.21. For the matrices in Example 2.20, find 2A, (−1)B, 21 C, 0C, 2A − B.
Notation: (−1)A is usually written as −A. 50
The rules for manipulating m × n matrices under the operations of addition and scalar multiplication are similar to those for numbers. This is not surprising because in performing any matrix operation we are just applying the same operation to the numbers in each of the (i, j) positions. Thus we have these standard properties: Matrix addition and scalar multiplication For all m × n matrices A, B, C, 0 (the zero matrix) and for all scalars λ, µ: (i) A + (B + C) = (A + B) + C
[Associative law]
(ii) A + B = B + A
[Commutative law]
(iii) (λ + µ)A = λA + µA
[Scalar laws]
(iv) λ(A + B) = λA + λB (v) λ(µA) = (λµ)A (vi) A + 0 = A
[Zero laws]
(vii) A − A = 0 We can now do some standard algebra with matrices: Example 2.22. If A, B and X are m × n matrices with 3A + 2X − B = 5A − X, find X in terms of matrices A and B.
Multiplication The product AB of matrices A and B can only be formed if the number of columns in A equals the number of rows in B. Then if A is an m × n matrix and B is an n × p matrix, the product C = AB is the m × p matrix with (i, j) entry given by cij = ai1 b1j + ai2 b2j + . . . + ain bnj =
n X
aik bkj .
k=1
That is: cij is found by multiplying together the corresponding entries from row i of A and column j of B, and then adding up the resulting products. A convenient way to check whether the product C = AB is defined, and if so to find its size, is to write down the size of A first then the size of B as shown. If the two inside numbers agree then the product is defined, and the two outside numbers give the size of C. 51
Example 2.23. If A is a 2 × 3 matrix, B a 3 × 1 matrix and C a 3 × 2 matrix, find the sizes of AB, BA, AC, CA, CB and C T B.
!
!
−3 9 2 3 Example 2.24. If A = and B = , find AB, BA and A2 . 2 −6 4 6
1 0 2 0 −1 Example 2.25. If A = and B = 2 1 , find AB and BA. 1 2 3 3 −1 !
52
Warning: The previous examples show that matrix multiplication differs from multiplication of numbers in certain ways. (1) For real numbers a and b, we always have ab = ba (called the commutative law for multiplication). However for matrices A and B, AB need not be equal to BA (and we say matrix multiplication is not commutative). Equality can fail to hold for 3 reasons: (i) AB may be defined whereas BA is not (see Example 2.23). (ii) AB and BA may both be defined but of different sizes (see Example 2.25). (iii) AB and BA may be defined and the same size but AB 6= BA (see Ex. 2.24). (2) For real numbers a and b, if ab = 0 then either a = 0 or b = 0. However the matrix equation AB = 0 (the zero matrix of the appropriate size) does not necessarily imply A = 0 or B = 0 (see Example 2.24). However the following standard multiplication properties hold as usual: Results about matrix multiplication For all matrices A, B, C for which the products are defined and for all scalars λ (i) A(BC) = (AB)C (ii) A(B + C) = AB + AC (A + B)C = AC + BC (iii) (λA)B = λ(AB) = A(λB).
[Associative Law] [Distributive Laws] [Scalar Law]
Remarks: Although both matrix addition and multiplication were defined for pairs of matrices, the associative laws on page 51 and above enable us to denote sums and products of three matrices as A + B + C and ABC without inserting any brackets. For no matter how brackets are inserted, the associative laws guarantee the same end result will be obtained. Because of the warning above, some of the familiar rules of factorization are no longer valid for matrices: Example 2.26. If a and b are numbers then (a + b)2 = a2 + 2ab + b2 . If A and B are square matrices find (A + B)2 .
53
A result concerning the transpose For all matrices A and B (for which multiplication is defined): (AB)T = B T AT Justification: Consider A =
(Note the reversed order!)
1 5 1 0 ,B= : 0 −1 2 1
Linear systems We now return briefly to systems of linear equations. Any linear system can be rewritten using a matrix product as shown in this example: Example 2.27. Write the following system of 3 equations in 3 unknowns as a simple matrix equation. x + 2y + 3z = 4 3x + y + 2z = 0 2x + 3y + z = 11
This example can be generalized in an obvious way. The system of m linear equations in n unknowns a11 x1 + a12 x2 + . . . + a1n xn = b1 a21 x1 + a22 x2 + . . . + a2n xn = b2 .. . am1 x1 + am2 x2 + . . . + amn xn = bm can be written as the matrix equation a11 a12 a13 · · · a1n x1 b1 a21 a22 a23 · · · a2n x2 b2 .. .. .. .. .. .. = .. . . . . . . . am1 am2 am3 · · · amn xn bm which has the form AX = B, where A X B
= matrix of coefficients, = column vector of unknowns, = column vector of constants.
Thus the problem of solving a system of linear equations reduces to solving the single matrix equation AX = B, where A and B are known and X is to be found. 54
We have already seen one way of solving this system: Gauss reduction on the augmented matrix (A | B). Another way involves using the inverse of the coefficient matrix.
2.9
Identity and inverse matrices
Definition. An identity matrix is a square matrix with 1s down the main diagonal of the matrix (where i = j) and 0s everywhere else. The n × n identity matrix is denoted In or just I when the size is obvious. So:
I2 =
I3 =
I4 =
The important property of the identity matrix In is that, when it is multiplied by a matrix A (of consistent size), the result is A itself. 1 4 5 Example 2.28. Find I2 A and AI3 where A = . 0 6 −1
Definition. The inverse of a square n × n matrix A is another n × n matrix denoted A−1 which has the property that AA−1 = A−1 A = In . 2 −1 2 1 Example 2.29. Show that B = is the inverse of A = . −3 2 3 2
Note. Not all matrices have inverses. Those that do are called invertible or nonsingular, that do not have inverses are called singular. For example, the while those 6 −3 matrix is singular: −8 4
55
A system of linear equations expressed in matrix form AX = B, can be easily solved if we know the inverse of the coefficient matrix A. Because then we can multiply both sides of the equation by this inverse:
Example 2.30. Given that the inverse of equations:
3 4 1 2
1 is 2
2 −4 , solve the system of −1 3
3x + 4y = 17 x + 2y = 5
3 −1 2 5 −4 −3 1 , solve the Example 2.31. Given that the inverse of 2 1 1 is −2 2 2 −3 2 −8 7 5 3x − y + 2z = 5 system of equations: 2x + y + z = 5 2x − 3y + 2z = 1
The inverse of a 2 × 2 matrix The inverse of a general invertible n × n matrix can be found by a method similar to Gauss reduction. (This is covered in MATH170.) The inverse of an invertible 2 × 2 matrix can be written down immediately using this formula: 1 a b d −b −1 If A = (where ad 6= bc) then its inverse is A = . c d ad − bc −c a
(If ad = bc then the matrix is not invertible.)
56
3 4 4 2 6 −3 Example 2.32. Find the inverses of A = ,B= ,C= . 1 2 −8 1 −8 4
2.10
Applications of matrices
Matrices are very similar to tables of numbers: they can be used to store all sorts of numerical information. Matrix addition and multiplication can often be used to combine or otherwise process information from different tables. Scenario 1 Suppose a manufacturer produces four items a, b, c, d. The number of items sold in one particular month is given by the sales matrix S and the price per unit for the items is given by the price matrix P , where: $20 $12 S = (30 25 50 12) P = $4 $10 For the first item, 30 are sold at $20 each so 30 × $20 = $600 is received from those sales. Similarly 25 × $12 = $300, 50 × $4 = $200, 12 × $10 = $120 are received from sales of the other three items. The total value of all sales is thus 30 × $20 + 25 × $12 + 50 × $4 + 12 × $10 = $1220. Notice that this calculation corresponds to the matrix product SP which produces a 1 × 1 matrix: (1220). Does the matrix product P S give any useful information?
57
Suppose that the matrix S below gives the sales of the four items (a, b, c, d) for four separate months, and the matrix M gives the profit and taxes (in $) on each item sold: c d a b Jan Feb S= Mar Apr
8 9 10 9 10 10 5 8 8 6 10 12 a
M=
Profit Taxes
b
6 5 3 5 c
d
200 250 100 50 80 150 30 30
!
58
Scenario 2 Suppose that the scores of the Highlanders’ players are kept in a table/matrix for each game. (Assume just 5 players A, . . . , E are involved for the sake of simplicity.) Game 1: A B G1 = C D E
try conv pen drop 0 3 2 0 2 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0
Game 2 : A B G2 = C D E
try conv pen drop 0 4 1 0 1 0 0 0 2 0 0 0 1 0 0 1 2 0 2 0
We could have similar matrices G3 , G4 , . . . ..., G18 , to cover all 18 Super Rugby games. For each player, we can calculate their total scores by adding these 18 matrices:
G1 + G2 + . . . + G18
A B =G= C D E
try conv pen drop 5 14 18 2 12 1 4 0 4 0 0 2 8 0 0 1 3 1 3 0
points try 5 conv 2 . Each score is awarded points given by the matrix P = pen 3 drop 3
59
Transition matrices Suppose that each year 90% of the people who live in towns stay there, while 10% move to the country. Suppose that 80% of the people who live in the country stay there, while 20% move to towns. We can store this information in a transition matrix (we use proportions rather than percentages): this year t c 0.9 0.2 t next T = 0.1 0.8 c year 70 If the column matrix P0 = shows the percentages of people living in town and 30 country at year 0, then what are the percentages at year 1, year 2, . . . , eventually?
60
Example 2.33. The change in weather from one day to the next in Dunedin is summarised in the transition matrix: sun 0.8 T = 0.1 0.1
first day cloud rain 0.2 0.2 0.7 0.3 0.1 0.5
sun cloud next day rain
(i) If the probabilities of sun/cloud/rain for next Monday are given in the matrix 0.6 PMon = 0.3, find the probability that it will rain on Wednesday. 0.1
(ii) Find the long-term proportions of weather types in Dunedin.
61
3 3.1
Complex numbers Introduction
We know from our experience of graphing quadratics of the form y = ax2 + bx + c, where a, b, c ∈ R, that they do not always have real roots. For example, there is no real number x, satisfying the quadratic equation x2 = −4,
(3.1)
since the square of any real number cannot be negative. Similarly x2 + 6x + 14 = 0 has no real solution, since by completing the square we may express this equation in the form (x + 3)2 = −5,
(3.2)
and again there is no real number whose square is −5. In this chapter we will consider a set of numbers called complex numbers and denoted C. These numbers contain the real numbers and provide solutions to all quadratic equations (including those above). In fact, they also provide solutions to all cubic, quartic — or in general to any “polynomial” — equations. The key to complex numbers is the introduction of the imaginary number i (also called the imaginary unit) which equals the square-root of −1: √ i = −1 so i2 = −1 With this new number, solve equations (3.1) and (3.2) above. For (3.1), x=
For (3.2),
2i, −2i, −3 + general form
√
5i and −3 −
√ 5i are examples of complex numbers, which have the
x + yi,
where
x, y ∈ R.
Other examples of complex numbers: 2 + 8i, −1 − i, 7i, 20. Note that a real number like 20 can be regarded as a complex number : 20 = 20 + 0i. Thus we say that the real numbers R form a subset of the complex numbers C.
62
3.2
Real and imaginary parts of z = x + yi
• x is called the real part of z, x = Re(z) • y is called the imaginary part of z, y = Im(z) • If Im(z) = 0 then z is a real number. As opposed to this, we will often call a complex number non-real if Im(z) 6= 0. • If Re(z) = 0 then z is a pure imaginary number.
3.3
Geometric representation: the complex plane
We can visualize real numbers as points on the “number line”.
How do we represent complex numbers? The complex number z = x + yi is formed from the ordered pair of real numbers (x, y), so it is natural to represent z by the point (x, y) in the usual coordinate plane:
In this context the plane is called the complex plane (or sometimes the Argand diagram). The x-axis is called the real axis, and the y-axis is called the imaginary axis. When a complex number is written in the form z = x + yi we say it is expressed in rectangular form.
3.4
The complex conjugate and modulus
The complex conjugate of z = x + yi is denoted z¯ and is the complex number z¯ = x − yi. It is obtained by negating the imaginary part of z. The modulus (or absolute value) of a complex number z = x + yi is denoted |z| and is defined as |z| =
p x2 + y 2 . 63
The complex conjugate of a complex number is the mirror image (or reflection) in the real axis. The modulus of a complex number is its distance from the origin.
3.5
Adding and multiplying
In practice, we manipulate complex numbers according to the familiar rules of realnumber algebra, replacing i2 wherever it occurs by −1. Adding and subtracting (2 + 3i) + (−1 + 4i) = (2 + 3i) − (−1 + 4i) = Multiplying 3(2 − 5i) = (2 + 3i)(−1 + 4i) = i(−1 + 2i) = (2 + 3i)(2 − 3i) = (3 + 4i)(4 + 3i) = (1 + i)4 =
(We will consider division later.) 64
Complex numbers are added and subtracted using the parallelogram law just as for vectors in the plane.
If z = x + yi and w = u + vi are complex numbers, then z + w = (x + u) + (y + v)i and z − w = (x − u) + (y − v)i can be represented geometrically.
3.6
Real and imaginary parts
Example 3.1. If z = 2 + 3i and w = 5 − 2i, then:
Now
Re(z) =
Im(z) =
Re(w) =
Im(w) =
Re(z + w) =
Im(z + w) =
z+w =
so: Note that: If z, w ∈ C then
Re(z + w) = Re(z) + Re(w), Im(z + w) = Im(z) + Im(w).
Example 3.2. Draw the following sets in the complex plane where z = x + yi: (Note: It is customary to draw a dotted line when a boundary line is not included in a set and a continuous line when it is.) A = {z ∈ C : Re(z) = 3}
B = {z ∈ C : Im(z) ≥ −2}
65
C = {z ∈ C : Re(z) = Im(z)}
3.7
D = {z ∈ C : Re(z + 2) < 3}
Solving equations involving complex numbers
Some equations involving a complex number z can best be solved by writing z = x + yi, substituting into the equation, and then equating the real and imaginary parts on both sides of the equation Example 3.3. Find all complex numbers z that satisfy 2z + i¯ z = 12 + 3i.
Example 3.4. Find
√
i.
66
Example 3.5. Find z such that z 2 = 5 − 12i.
(Here, we have found the square roots of 5 − 12i. Later, we will consider a more general approach to do this.) The quadratic formula: 2
ax + bx + c = 0
⇒
x=
−b ±
√
b2 − 4ac 2a
can be applied to equations with complex coefficients a, b and c. Example 3.6. Use the quadratic formula to solve the quadratic equation z 2 − iz + 2 = 0
67
3.8
Conjugates
Example 3.7. If z = 2 + 3i and w = 1 − i, then: z¯ = z + z¯ = z − z¯ = z¯ + w ¯ = z+w = z¯w¯ = zw = In fact If z, w ∈ C, then z¯ = z,
z + z¯ = 2 Re(z), z + w = z¯ + w, ¯
3.9
z − z¯ = 2 Im(z)i, zw = z¯w. ¯
Modulus
Example 3.8. If u = −2, v = 3i, w = 3 + 2i, z = 5 − i, then: |u| =
|v| =
|w| =
|z| =
|¯ z| = z z¯ = |zw| =
In fact: If z, w ∈ C, then |¯ z | = |z|,
z z¯ = |z|2 ,
|zw| = |z||w|.
68
The next result gives a very important and useful interpretation of the modulus: Result. The distance between the two complex numbers w and z in the complex plane is given by |z − w|. Proof.
Example 3.9. Interpret |2i − 3| in two ways.
Example 3.10. Find the distance between the complex numbers z = −1 + i and w = 3 − 2i in the complex plane.
69
Example 3.11. Draw the following sets in the complex plane : A = {z ∈ C : |z − (3 + i)| = 2}
B = {z ∈ C : |z + i| ≤ 2}
C = {z ∈ C : 1 < |z + i| ≤ 2}
D = {z ∈ C : |z − i| = |z + i|}
3.10
Division
We first consider the reciprocal of a complex number. 1 1 as a complex number in standard For example, if z = 3 + 2i, can we express = z 3 + 2i rectangular form? Recalling that z z¯ = |z|2 , if we multiply top and bottom by the complex conjugate z¯ = 3 − 2i, then we get a real number in the denominator: 1 3 − 2i 3 − 2i 3 2 1 = = = − i 3 + 2i 3 + 2i 3 − 2i 13 13 13 We have thus expressed
1 3 2 in the rectangular form − i. 3 + 2i 13 13 70
1 in rectangular form. 2−i
Example 3.12. Find
z¯ z¯ 1 = = 2 So we have that (assuming z 6= 0): z z z¯ |z| 1 w¯ z w¯ z w =w = = 2 We can now define division: z z z z¯ |z| 1+i in rectangular form. 2−i
Example 3.13. Find
3 + 5i 3 + 5i . Example 3.14. Find in rectangular form, and 4+i 4+i
As illustrated in the last example:
w |w| = z |z|
Summary of key definitions and results Rectangular form: z = x + yi Real part:
Re(z) = x
Imaginary part:
Im(z) = y
|¯ z | = |z|,
z z¯ = |z|2 , z + w = z¯ + w, ¯
Complex conjugate: z¯ = x − yi p Modulus: |z| = x2 + y 2 |zw| = |z||w|,
w |w| = z |z|
zw = z¯w¯
71
3.11
The Mandelbrot set
Suppose we take a complex number w, and define a sequence of complex numbers {zn } as follows: z0 = w,
z1 = z02 + w,
z2 = z12 + w,
z3 = z22 + w,
etc.
What happens to the sequence {zn } as n → ∞? Do the numbers get larger and larger in modulus without bound, or do they converge in on a single number, or do they have some other behaviour? In fact we can prove that if any value of zn is outside the circular region |z| ≤ 2, then all subsequent values will be outside that region and they get larger and larger in modulus. In 1979 the Polish mathematician Benoˆıt Mandelbrot (working at IBM) considered for which values of w the corresponding sequence did not escape this circular region. The set of all such numbers is now known as the Mandelbrot set. If {zn } never escapes from the circular region |z| < 2 then we say that the original number w is in the Mandelbrot set. Such numbers w are usually coloured black in pictures of this set. If {zn } does eventually escape from the circular region |z| < 2 then we say w is not in the Mandelbrot set. Such numbers w can be coloured according to the value of n of the first number zn which is outside the circle. 2 2 Example 3.15. Determine whether the complex numbers, 1, −1, i, + i are in the 3 3 Mandelbrot set. The successive z values starting with w = 1 are: z = 1, 2, 5, . . . so the sequence does escape, and so 1 is not in the Mandelbrot set. The successive z values starting with w = −1 are: z = −1, 0, −1, 0, . . . so the sequence is trapped, and so −1 is in the Mandelbrot set.
72
3.12
Argument
The argument of a non-zero complex number z, denoted arg z , is the angle between the positive real axis and the line segment from O to z, measured positively in an anticlockwise direction. Since the same angle can be represented in different ways (for example, −5π/4, 3π/4, 11π/4 all denote the same angle), the argument of a complex number is determined only up to multiples of 2π (or 360◦ ). To avoid this ambiguity we will usually restrict the range of arg z to the interval (−π, π]. The arguments of some complex numbers will be obvious from a quick sketch:
If z = x + yi, then θ = arg z can be obtained from: y y or θ = tan−1 tan θ = x x
But we must be careful since the trigonometric equation tan θ = a has two solutions in the interval (−π, π]. Put another way, the arguments of the complex numbers x + yi and −x − yi are different and yet they have the same tangents.
We must take care to select the correct angle — drawing z in the complex plane will help to ensure we do this. 73
To determine the argument of z , θ = arg z, where z = x + yi:
If x = 0:
If x 6= 0: −1 y (i) Find the angle α = tan . x π Calculators will give a value between 0 and . 2 (ii) Sketch the position of z in the complex plane, and choose the relevant θ:
Example 3.16. Find θ = arg z for the following complex numbers: (i) 1 + 2i
√ (ii) − 3 − i
(iii) −2 + i
(iv) 3 − 2i
74
3.13
Polar forms (modulus-argument forms)
Instead of using the standard rectangular form z = x + yi, the complex number z may be completely specified by its modulus and argument. For θ = arg z describes in which direction (from the origin) to go, and |z| how far. Note that x = |z| cos θ and y = |z| sin θ. A complex number which is given in the form z = |z|(cos θ + i sin θ) where θ = arg z, is said to be in polar form – we can see immediately what the modulus and argument are so we also call this the modulus-argument form. For convenience, we abbreviate the complex number cos θ + i sin θ by cis θ, cis θ = cos θ + i sin θ So: cis π = π = cis − 3 and we have: z = r cis θ,
where
r = |z|,
θ = arg z
using this cis notation in the polar form. There is an equivalent Euler or exponential notation eiθ which comes about as follows: Let
z = cis θ = cos θ + i sin θ,
then
iz =
But also
dz = dθ
Solving this differential equation (refer to Calculus lectures) gives the general solution z = A eiθ , for some constant A. Now z = 1 when θ = 0, so this gives that A = 1 and hence z = eiθ . We thus have: cis θ = cos θ + i sin θ = eiθ and so: z = r eiθ ,
where r = |z|,
θ = arg z
75
Example 3.17. Express the following complex numbers in both cis and Euler polar forms: (i) −3i
(ii) 1 +
√
3i
(iii) −1 + i
Of course when z is given in polar form one can easily read off its modulus and argument. Example 3.18. Find the modulus and argument of z = 2e7π/3 i
To change a complex number from polar form to rectangular form it helps to know the sines and cosines of common angles, together with such identities as cos(−θ) = cos θ, sin(−θ) = − sin θ cos(π − θ) = − cos θ, sin(π − θ) = sin θ Example 3.19. Express the following complex numbers in rectangular form. (i) z = cis(π/2)
(ii) w = 2e3π/4i
(iii) v = 6 cis(11π/6)
76
(iv) u = 10e3i
(We will concentrate on the Euler notation from here on.)
3.14
Multiplication and division of polar forms
Multiplication:
So, to multiply z and w, multiply their moduli and add their arguments. |zw| = |z||w| arg(zw) = arg(z) + arg(w) (mod 2π)
π
π
Example 3.20. If z = 2e 2 i , w = 3e 4 i and v = 1 − i, find zw and zv.
π
π
Example 3.21. If z = 2e 3 i and w = 4e 6 i , find zw in rectangular form.
77
We may also interpret multiplication geometrically as follows: The effect of multiplying z by w is to (i) rotate anti-clockwise by ϕ = arg w (ii) scale the modulus by a factor |w| Example 3.22. Describe geometrically the effect of multiplying any complex number z by (i) i
(ii) 1 + i
Example 3.23. For z = eiθ shown, draw z 2 , z 3 , z 4 , . . .
z θ
le
it un
rc ci
78
Division:
So, to divide z by w (6= 0), divide their moduli and subtract their arguments. z |z| = w |w| arg(z/w) = arg(z) − arg(w) (mod 2π) π
Example 3.24. If z = 1 + i and w = 2e 3 i , find
3.15
z w
in polar form.
De Moivre’s theorem (cis θ)n = cis nθ,
for any natural number n
This result, known as de Moivre’s theorem, says that in order to raise a complex number on the unit circle to the nth power, all that we need to do is stay on the unit circle and multiply the argument by n. We can see how this works when n = 2:
The result can be proved easily for general n by using a technique called “proof by induction”. However, using the Euler form we see that De Moivre’s Theorem follows immediately: (cis θ)n = (eiθ )n = einθ = cis nθ
79
2π
Example 3.25. If z = e 3 i , find z 2 and z 3 in polar form.
Example 3.26. If z = 1 +
3.16
√
3i, find z 2 , z 3 , z −1 and z −2 in polar form.
Roots of a complex number
If w is a complex number and n is a natural number, then any number z such that z n = w is called an nth root of w. For example, 1 +
√ √ 3i is a cube root of −8 since (1 + 3i)3 = −8.
The roots of “unity” Although the equation xn = 1 has only 2 possible real solutions, x = 1 and x = −1 (the first occurring always but the latter only when n is even), it turns out that: There are always n distinct complex numbers z such that z n = 1. Example 3.27. Find the cube roots of 1.
80
Result. The nth roots of 1 are e(2πk/n)i ,
k = 0, 1, 2, . . . , n − 1.
These roots are equally spaced around the unit circle at angles 2π/n apart, starting at 1.
General roots Example 3.28. Find the 5th roots of w =
√ 3 + i.
Result. The nth roots of w = reiϕ are 1
rne
ϕ+2πk i n
,
k = 0, 1, . . . , n − 1.
These roots are equally spaced around a circle of radius r1/n at an angle 2π/n apart, starting at the angle ϕ/n. Example 3.29. Find the square roots of 1 +
√
3i.
81
Example 3.30. Solve the equation z 6 = 27i.
Example 3.31. Find the cube roots of 5 + 12i to 3 decimal places.
Example 3.32. Find the 4th roots of 16 cis 100◦ .
Example 3.33. Solve z 5 = −2i (use degrees for the angles).
82
4 4.1
Polynomials Introduction
A polynomial function can be written as a sum of terms of the form axj where a is a constant (called a coefficient), j is a non-negative integer (the power), and x is the variable. Thus in general a polynomial can be expressed as p(x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 which is called its power form. Examples of polynomials:
The degree of a polynomial is the highest of all the powers of the variable. The leading coefficient is the coefficient of the term with the highest power. The constant term is the term without x (i.e. with a power of 0).
p(x) = 3x4 leading coefficient
4x3 + 7x 30
degree
constant term
p(x) = 2 + x3 2x5 + 6x2
10x
A monic polynomial is one with leading coefficient 1. Polynomials are manipulated using familiar arithmetic rules for addition and multiplication. For example, 3(2x2 + x − 2) = 6x2 + 3x − 6, (x + 1)(x2 − x + 1) = x3 − x2 + x + x2 − x + 1 = x3 + 1.
83
4.2
Sketching polynomials
For the polynomial y = p(x) we can say the following about the way the graph behaves for large positive or negative values of x: • If the leading coefficient is positive then as x → +∞, y → +∞.
If the degree is even then as x → −∞, y → +∞.
If the degree is odd then as x → −∞, y → −∞.
• If the leading coefficient is negative then we need to negate the above limits for y.
• At any single real root, the graph crosses the x-axis.
• At any double real root, the graph just touches the x-axis.
• At any triple real root, the graph crosses the x-axis, but at the intersection point the slope is zero.
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4.3
Dividing one polynomial by another
When we divide one polynomial by another it is unlikely that the result will be a polynomial. For example, when we divide x2 − 3x + 2 by x + 1, we get the rational x2 − 3x − 2 , not a polynomial. function, x+1 This mirrors the situation with integers: when we divide one integer by another we are unlikely to get an integer as a result. For example, when we divide 20 by 7 we get the 20 ratio, or fraction, . However we can write 20 = 2 × 7 + 6 and say that when we divide 7 20 by 7 we get a quotient of 2 and a remainder of 6. Similarly, dividing x2 − 3x + 2 by x + 1 gives a quotient of x − 4 and a remainder of 6. That is, we can write:
Result 1 — Quotient and remainder Given polynomials p(x) and s(x) with the degree of s less than or equal to that of p, there are polynomials q(x) (the quotient) and r(x) (the remainder) such that p(x) = q(x)s(x) + r(x) with the degree of r less than that of s. Finding quotient and remainder — general case One polynomial can be divided by another using a process of long division similar to that used for dividing numbers (before calculators!!). Two examples will illustrate the method: Example 4.1. Find the quotient and remainder when p(x) = x3 − 5x2 + 6 is divided by s(x) = x + 2.
85
Example 4.2. Find the quotient and remainder when p(x) = 3x4 + x3 − 2x + 4 is divided by s(x) = x2 + 2x − 1.
Finding quotient and remainder — special case When we are dividing by the linear polynomial x − α, the remainder will be a constant. There is a quicker way of finding both quotient and remainder than the long division method. This is called. . .
4.4
Synthetic division
Example 4.3. Find the quotient and remainder when p(x) = 4x4 − 7x3 − 20x − 25 is divided by x − 3.
Example 4.4. Find the quotient and remainder when p(x) = x3 − 5x2 + 6 is divided by x + 2. (Same as Example 4.1)
86
Let’s see how synthetic division works. Suppose we wish to divide the polynomial p(x) by x − α. Then p(x) = (x − α)q(x) + r.
(4.1)
Now if p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0 and q(x) = bn−1 xn−1 + . . . + b1 x + b0 and we equate the coefficients on both sides of (4.1), then we can show that: an = bn−1 an−1 = bn−2 − αbn−1 an−2 = bn−3 − αbn−2 .. . a1 a0
= b0 − αb1 = r − αb0
so:
bn−1 = an bn−2 = an−1 + αbn−1 bn−3 = an−2 + αbn−2 .. . b0 r
= a1 + αb1 = a0 + αb0
These quantities — i.e. the coefficients of q and the remainder r — are the column sums in the synthetic division table.
4.5
The remainder theorem
From (4.1) above we see that p(α) = (α − α)q(α) + r = r. Therefore: Result 2 — The Remainder Theorem When p(x) is divided by x − α, the remainder is p(α). Example 4.5. Find the remainder when the polynomial p(x) = 2x3 − 4x2 − 7x + 5 is divided by x − 3.
Example 4.6. The polynomial p(x) = ax3 + 2x2 − 2x − 6 gives a remainder of 3 when divided by x + 1. Find a.
87
Example 4.7. The polynomial p(x) = x3 + ax2 − 5x + b gives a remainder of 1 when divided by x − 2 and a remainder of 2 when divided by x − 1. Find a and b.
4.6
Factors and roots
The polynomial s(x) is a factor of the polynomial p(x) if we can write p(x) = q(x)s(x) for some other polynomial q(x).
α is a root (or zero) of the polynomial p(x) if p(α) = 0. These are connected by: Result 3 — The Factor Theorem x − α is a factor of p(x) if and only if α is a root of p(x).
Examples: • The polynomial p(x) = x3 + 4x2 − 9 has x + 3 as a factor since we can write x3 + 4x2 − 9 = (x + 3)(x2 + x − 3) It is clear that −3 is a root of p since p(−3) = −27 + 36 − 9 = 0. • For p(x) = x4 − 6x3 + 22x + 15, we find that p(5) =
Result — “Fundamental theorem of algebra” Every polynomial p(x) of degree n > 0 has at least one root (in general this is complex, but it could be real). From this we conclude: Every polynomial p(x) of degree n > 0 has exactly n roots (real or complex). (Some of the roots may be repeated.) 88
We may be able to factorize a polynomial completely in terms of linear factors, from which its roots are immediately obvious. • The polynomial p(x) = 6x4 + 13x3 − 8x2 − 17x + 6 can be written p(x) = (2x + 3)(x + 2)(3x − 1)(x − 1) Hence its roots are • The polynomial p(x) = x4 − 3x3 + 3x2 − 3x + 2 can be written p(x) = (x + i)(x − i)(x − 1)(x − 2) Hence its roots are
4.7
Real polynomials
A real polynomial is one with real coefficients (the only type we have so far considered). These are not real polynomials: x2 + ix − 2,
2x3 + 3x − 1 + 4i.
We can always factorize a real polynomial into real quadratic factors together with a linear factor if its degree is odd. For example: p(x) = x4 − 2x3 − x2 − 16x + 10 = (x2 + 2x + 5)(x2 − 4x + 2) q(x) = x5 − 8x3 − 9x = (x − 3)(x2 + 1)(x2 + 3x) The quadratic factors may correspond to real or complex roots. • If the roots are real then the quadratic can be factorized further into real linear factors. • If the roots are non-real they occur as complex conjugate pairs.
The quadratics in the factorization of p(x) above lead to roots −1 ± 2i and 2 ± we can factorize this polynomial using these linear factors:
√ 2 . So
p(x) = The quadratics in the factorization of q(x) above lead to roots i, −i, 0 and −3. This polynomial can be factorized as: q(x) =
89
Example 4.8. Given that 1 − 3i is a root, find all roots of the polynomial p(x) = x3 + 6x + 20.
Example 4.9. Given that 1 + i is a root, find all roots of the polynomial p(x) = x4 − 3x3 + 2x2 + 2x − 4.
4.8
Rational roots of a real polynomial
Suppose that the polynomial p(x) = an xn + . . . + a0 has integer coefficients, and at least one real linear factor. Suppose this linear factor is (αx − β), where α and β are integers with no common factors. Then αβ is a root of p, (in fact a rational root) and so p αβ = 0. That is: n n−1 β β β an + an−1 + . . . + a1 + a0 = 0. α α α
(4.2)
Multiplying through (1.4) by αn and taking the last term to the right-hand side: 90
Now β is a factor of the left-hand side, so it must also be a factor of the right-hand side. But we know that β is not a factor of α, so it must be a factor of a0 . Also, isolating the term an β n we get: an β n = −an−1 β n−1 α − . . . − a1 βαn−1 − a0 αn . Here α is a factor of the right-hand side, so it must also be a factor of the left-hand side. But we know that α is not a factor of β, so it must be a factor of an . β is a (rational) root of p Thus α where β is a factor of the constant term of p, α is a factor of the leading coefficient of p. The consequence of this is that for a polynomial p with integer coefficients, there are only a limited number of rational numbers which could be roots. β The candidates for roots are ± , α where β is a factor of the constant term, and α is a factor of the leading coefficient. Example 4.10. Find the possible candidates for rational roots of the polynomial p(x) = 3x3 + 2x2 + 17x − 6.
Example 4.11. Find the possible candidates for rational roots of the polynomial p(x) = 2x6 + 2x4 − 200x − 77.
91
Warning: This result does not guarantee that the polynomial in question will have any rational roots. We can only list the candidates and then test to see if indeed they are roots. Example 4.12. Find all roots of the polynomial p(x) = 4x4 + 11x3 + x2 − 9x + 2.
4.9
Repeated roots
We say that α is a repeated root (or multiple root) of a polynomial p(x) if we can write p(x) = (x − α)n q(x) (4.3) (for some polynomial q(x)), where n > 1. In particular, if n = 2, then α is a double root, if n = 3, then α is a triple root. Differentiating (4.3) we get p0 (x) =
This proves the following convenient test for a polynomial to have a repeated root. Result — Repeated roots α is a repeated root of p(x) if and only if both p(α) = 0 and p0 (α) = 0. Example 4.13. Given that p(x) = 2x3 − 5x2 − 4x + 12 has a repeated root, find this root.
92
Example 4.14. The polynomial p(x) = 2x3 + ax + b has a repeated root at x = 3. Find a and b and factorize the polynomial completely.
4.10
Final examples
Example 4.15. Investigate the roots of the polynomial 4x3 − 2x2 − 9.
93
Example 4.16. Find the roots of the polynomial x4 − 2x3 − 3x2 + 2x + 2.
94
A
Some common algebraic mistakes
Incorrect
Correct
2(a + 3) = 2a + 3
2(a + 3) = 2a + 6
x − (a + 2) = x − a + 2
x − (a + 2) = x − a − 2
1 = 0; 0
Both
0 =1 0
1 0 and are undefined 0 0
(a + b)2 = a2 + b2
(a + b)2 = a2 + 2ab + b2
a2 a3 = a6
a2 a3 = a5
(a2 )3 = a5
(a2 )3 = a6
√
√
√ √ √ √
9 = ±3
√
4a = 2a
√
a2 = a a+b=
√
a+
√ b
a2 + b 2 = a + b
√ √
9=3 4a =
√ √ √ 4 a=2 a
a2 = |a| a + b cannot be simplified a2 + b2 cannot be simplified
u u u + = v w v+w
u(v + w) u u + = v w vw
u+w u w + = v x v+x
ux + vw u w + = v x vx
r ru = su + tv s + tv
ru cannot be simplified su + tv
sin(2x) = 2 sin x
sin(2x) = 2 sin x cos x
sin(x2 ) = (sin x)2
sin(x2 ) can’t be written in terms of sin x
2
2
ex = (ex )2
ex = (ex )x
ln(u + v) = ln u + ln v
ln(uv) = ln u + ln v
Many of these are obvious√if you try √ some √ easy values of the √ variables: for example no-one would believe that 1 + √ 1 = 1 + √ 1, as√one side is 2 and the other is 2, but many people would accept that a + b = a + b is a valid rule!
95
B
Basic trigonometry
Radian measure In a circle with radius 1, the angle θ in radians is equal to the arc length. So if the arc is 2 units long, the angle is 2 radians.
arc length
1
θ
angle
θ
Taking a full turn, the arc length equals the circumference (2π), so the angle in radians for a complete turn is also (2π): angle = arc length = 2π So 360◦ = 2π radians and consequently 180◦ = π radians.
angle θ in radians = arc length
π 180◦
So: to convert degrees to radians, multiply by
180◦ to convert radians to degrees, multiply by π 3 e.g. convert π to degrees. 4
3 3 180◦ π= π× = 135◦ 4 4 π
e.g. convert 120◦ to radians. 120◦ = 120 ×
π 2π = ◦ 180 3
................................................................................ Basic trig functions In a right-angled triangle we have these important functions of an angle: sin θ =
opp hpy
”soh”
cos θ =
adj hpy
”cah”
tan θ =
opp ”toa” adj
adj
SIDES: adjacent to θ = "adj" opposite θ = "opp" hypotenuse = "hyp"
hyp
θ
opp Note: tan θ =
sin θ cos θ
................................................................................
e.g. Find cos θ, θ, and the length of x 7 cos θ = ⇒ θ = cos−1 (7/9) = 0.68 9
9
x θ
7 x tan θ = so x = 7 tan(0.68) = 7 × 0.808 = 5.66 units 7 (or use Phythagoras’ theorem: x2 + 72 = 92 ) ................................................................................
96
Standard right angled triangles There are two standard right angled triangles which are used frequently in trigonometry and should be memorised.
sides are in the
π 3
1
sides are in the
π
ratio 1 : 3 : 2
2
1
π 6
Note:
3
π = 30 6
2
4
ratio 1 : 1 : 2
π
π = 60 3
1
4
Note:
π = 45 4
Examples 2 2
60
θ = 45
3
π
θ = 6 6 θ
h h = 4
θ
With the standard triangles at the top we have: √ π 1 π 3 sin = cos = 6 2 6 2 1 π =√ 4 2 √ π 3 sin = 3 2 sin
tan
π 1 =√ 6 3
cos
1 π =√ 4 2
tan
π =1 4
cos
π 1 = 3 2
tan
π √ = 3 3
3
................................................................................
1 θ O cos θ
x = cos θ, y = sin θ
sin θ
P(x, y)
Finding sin/cos/tan of other angles Let P be the point (x, y) on the circle, centre 0, radius 1. Then where θ is the angle measured anti-clockwise from the psotitive x-axis to the radius OP . This allows us to define these functions for any angle. Graphs:
sin x
Π 2
Π
3 Π 2
cos x
2Π
x
Π 2
Π
3 Π 2
tan x
2Π
x
- Π 2
Π 2
Π
3 Π 2
2Π x
97
Identities π π so ϕ = − θ. 2 2 Then cos θ = b/c = sin ϕ = sin(π/2 − θ) sin θ = a/c = cos ϕ = cos(π/2 − θ) 1 1 1 tan θ = a/b = = = b/a tan ϕ tan(π/2 − θ)
In this triangle θ + ϕ =
Also a2 + b2 = c2 using Pythagoras’ theorem. Dividing by c2 : or
b
c
θ
a sin2 θ + cos2 θ = 1 cos θ = sin(π/2 − θ) sin θ = cos(π/2 − θ) 1 tan θ = tan(π/2 − θ)
a2 b 2 + =1 c2 c2 sin2 θ + cos2 θ = 1
sin 2θ = 2 sin θ cos θ cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ Other identities and functions:
cos(−θ) = cos θ sin(−θ) = − sin θ tan(−θ) = − tan θ sec θ = 1/ cos θ
cos(π − θ) = − cos θ sin(π − θ) = sin θ tan(π − θ) = − tan θ cosec θ = 1/ sin θ
cot θ = 1/ tan θ
Examples (i) Find cos θ and sin 2θ given that sin θ = 0.8 and 0 < θ < π/2. p cos2 θ = 1 − sin2 θ so cos θ = + 1 − (0.8)2 = 0.6 We take the positive square root because, for the stated θ, cos θ is positive (see graph).
sin 2θ = 2 sin θ cos θ = 2(0.8)(0.6) = 9.6 (ii) Find cos(2π/3) and sin(−3π/4). cos(2π/3) = cos(π − π/3) = − cos(π/3) = −1/2 √ sin(−3π/4) = − sin(3π/4) = − sin(π − π/4) = − sin(π/4) = −1/ 2 (iii) Find sin θ given that sec θ = 4 and π < θ < 2π. cos θ = 1/ sec pθ = 0.25 so sin θ = − 1 − (0.25)2 = 0.9682
We take the negative square root because, for the stated θ, cos θ is negative (see graph).
(iv) A ladder 5m long rests against a wall. It makes a angle of 36◦ with the ground. How far is the bottom of the ladder from the wall? if x = distance required, then cos 36◦ = x/5m. so x = 5m cos 36◦ = 4.045m.
98
Index absolute value (complex number), 63 acceleration, 31 angle, 16 acute, 16 between two planes, 24 obtuse, 16 arc-length, 32 Argand diagram, 63 argument, 75 argument (complex number), 73 augmented matrix, 36
imaginary unit, 62 inconsistent system, 47 integer coefficients, 90 inverse matrix, 55 invertible matrix, 55
cis, 75 coefficient (polynomial), 83 collinear, 11 column matrix/vector, 49 columns, 48 complex conjugate, 63, 64 complex numbers, 62 complex plane, 63 computer graphics, 33 consistent system, 47 constant term (polynomial), 83 cosine rule, 14
magnitude, 1, 2, 8 Mandelbrot set, 72 matrix, 48 matrix addition, 51 matrix equation, 54 matrix multiplication, 51 modulus, 63, 64, 75 modulus-argument form, 75 monic polynomial, 83 multiple root, 92
de Moivre’s theorem, 79 degree (polynomial), 83 derivative of a vector, 31 direction (vector), 1, 2 division of complex numbers, 71 double real root, 84 double root, 92
orthogonal, 15
elements of a matrix, 48 entries of a matrix, 48 equivalent matrices, 39 Euler notation, 75 exponential notation, 75 factor of a polynomial, 88 free variables, 45 Gauss reduction, 39, 40 general solution, 45 identity matrix, 55 imaginary axis, 63 imaginary number, 62 imaginary part, 63
leading coefficient (polynomial), 83 leading entry (matrix row), 39 leading variables, 45 linear equation, 35 long division, 85
normal equation of a plane, 22 normal vector, 22
parallel planes, 27 parallel vectors, 10 parallelogram, 5 parametric equation of a line, 17 parametric equation of a plane, 29 perpendicular, 15 polar form, 75 polynomial, 83 position vector, 5 power (polynomial), 83 power form (polynomial), 83 projection, 20, 27, 33 projection vector, 21 quotient, 85 rational root, 90 real axis, 63 real numbers, 62 real part, 63 real polynomial, 89 99
rectangular form, 63 reduced row echelon form, 39 remainder, 85 remainder theorem, 87 repeated root, 92 root (complex number), 80 root (polynomial), 88 row matrix/vector, 49 rows, 48 scalar, 1, 50 scalar multiplication, 51 scalar product, 13 section formula, 11 sense (vector), 1, 2 single real root, 84 singular matrix, 55 size of a matrix, 48 solution set, 35 square matrix, 49 system of linear equations, 35 transition matrix, 60 transpose, 50 triangle law of vector addition, 3 triple real root, 84 triple root, 92 unit vector, 9 variable (polynomial), 83 vector, 1 2-dimensional, 4 3-dimensional, 4 vector product, 13 velocity, 31 view-plane, 33 zero (polynomial), 88 zero matrix, 49 zero vector, 2
100
