MAT01A1: Complex Numbers (Appendix H)

Transcription

1 MAT01A1: Complex Numbers (Appendix H) Dr Craig 13 February 2019

2 Introduction Who: Dr Craig What: Lecturer & course coordinator for MAT01A1 Where: C-Ring 508 Web:

3 Important information Course code: MAT01A1 NOT: MAT1A1E, MAT1A3E, MATE0A1, MAEB0A1, MAA00A1, MAT00A1, MAFT0A1 Learning Guide: available on Blackboard. Please check Blackboard twice a week. Student check this account twice per week or set up forwarding to an address that you check frequently.

4 Important information Lecture times: Tuesday 08h50 10h25 Wednesdays 17h10 18h45 Lecture venues: C-LES 102, C-LES 103 Tutorials: Tuesday afternoons 13h50 15h25: D-LES 104 or B-LES 102 OR 15h30 17h05: C-LES 203

5 Other announcements No tuts for MAT01A1 on Wednesdays. If you see this on your timetable, it is an error. (To move your Chem. prac., Mr Kgatshe CSC02A2 students. Dr Craig regarding tutorial clash. Maths Learning Centre in C-Ring 512: 10h30 15h25 Mondays 08h00 15h25 Tuesday to Thursday 08h00 12h05 Fridays

6 Lecturers Consultation Hours Monday: 14h40 15h25 Dr Craig (C-508) Tuesday: 11h20 13h45 Dr Robinson (C-514) Wednesday: 15h30 17h05 Dr Robinson (C-514) Thursday: 11h20 12h55 Dr Craig (C-508) Friday: 11h20 12h55 Dr Craig (C-508)

7 Saturday class 09h00 12h00 C-LES 401 Questions will be from all topics covered so far. Please attend if you want to spend extra time on your maths.

8 Warm up Example: Find all of the values of x in the interval [0, 2π] such that sin x = sin 2x. Example: For x [0, 2π], solve 1 < tan x < 1 To do this we will first solve tan x = 1 and tan x = 1 and then use the graph of y = tan x.

9 Today s lecture Complex numbers: introduction and basic operations Polar form (including multiplication and division) Powers of complex numbers (De Moivre s Theorem) Why do we cover complex numbers? Applications in Physics and Electronics. Also, studied in mathematics in Complex Analysis (3rd year module).

10 Complex numbers A complex number is made up of a real part and an imaginary part. The imaginary part involves the square root of 1. We define i = 1. All complex numbers will be of the form: z = a + bi where a, b R and i 2 = 1 The set of complex numbers is denoted by C. (Remember, N is the natural numbers, Z is the integers, R is the real numbers.)

11 Complex numbers in the plane The complex number z = a + bi is represented on the complex plane by the point (a, b). For z = 2 + i we have a = 2 and b = 1. Im (a, b) = (2, 1) R

12 More numbers in the complex plane 4 + 2i Im 2 + 3i R 2 2i 3 2i

13 Addition and subtraction in C To add/subtract complex numbers, we simply add/subtract the real and imaginary parts separately and then combine them. Let z = a + bi and w = c + di. Then Subtraction: z + w = (a + bi) + (c + di) = (a + c) + (b + d)i z w = (a + bi) (c + di) = (a c) + (b d)i

14 Multiplication Let z = a + bi and w = c + di. Then z w = (a + bi)(c + di) = ac + adi + bci + (bi)(di) = ac + (ad + bc)i + bd(i 2 ) = ac + (ad + bc)i + ( 1)(bd) = (ac bd) + (ad + bc)i The real part of z w is ac bd and the imaginary part is ad + bc.

15 Examples Add: z = 2 7i and w = 4 + 2i. Calculate: (1 + i) (3 4i). Multiply: z = 2 + 3i and w = 4 2i.

16 The conjugate of a complex number Consider the complex number z = a + bi. The complex conjugate of z is the complex number z = a bi Im z z (a, b) R (a, b)

17 Division of complex numbers To divide complex numbers we make use of the complex conjugate of the denominator. Let z = a + bi and w = c + di. Then z w = a + bi c + di = a + bi c + di c di c di = (a + bi)(c di) c 2 + d 2 = ac + bd c 2 + d 2 + bc ad c 2 + d 2 i

18 Now that we know how to divide, we can consider reciprocals of complex numbers: 1 z = 1. z z. z = a bi ( ) ( ) a b a 2 + b = + i 2 a 2 + b 2 a 2 + b 2

19 Conjugates and absolute value Properties of conjugates: z + w = z + w zw = z w z n = z n The absolute value, or modulus, of a complex number is the distance from the origin in the complex plane. If z = a + bi then z = a 2 + b 2 We see that z. z = z 2

20 Roots of quadratic equations in C ax 2 + bx + c = 0 x = b ± b 2 4ac 2a When we allow complex roots as solutions, we can apply the above formula to cases when b 2 4ac < 0. When y 0, we let y = ( y)i. Thus every quadratic has complex roots. Example: solve x 2 + x + 1 = 0 for x C

21 Complex roots of polynomials Consider a polynomial of degree n with coefficients from R. Such a polynomial has the general form a n x n + a n 1 x n a 1 x + a 0 A root of a polynomial is a value of x that makes the polynomial equal to zero. Every polynomial of degree n has n complex roots.

22 Complex roots of polynomials Every polynomial of degree n has n complex roots. (Roots might be repeated, e.g. x 2 = 0.)

23 Terminology: A complex number z is said to be in rectangular form when it is written as z = a + bi. This terminology distinguishes rectangular form from the one we are about to introduce: polar form.

24 Polar form Any complex number z = a + bi can be considered as a point (a, b). Thus it can also be represented by polar coordinates as (r, θ). Im (a, b) r b θ a R Now a = r cos θ and b = r sin θ.

25 Since a = r cos θ and b = r sin θ, any complex number z = a + bi can be written as z = r(cos θ + i sin θ) where r = z = a 2 + b 2 and tan θ = b a. The angle θ is called the argument of the complex number z. We write θ = arg(z). Note: arg(z) is not unique. If θ = arg(z) then we also have θ + k.2π = arg(z) where k is any integer (k Z).

26 Finding the argument of z C When converting a complex number into polar form the best method for finding the argument of z = a + bi is to plot z. If you use the fact that tan θ = b then there a are two possible solutions for θ [0, 2π].

27 Examples: z = 1 + 3i and w = 1 i. Find z, arg(z), w, arg(w). ( 1, 3) Im r R (1, 1)

28 Multiplication and division in polar form Let z 1 = r 1 (cos θ 1 + i sin θ 1 ) and z 2 = r 2 (cos θ 2 + i sin θ 2 ). We use the addition and subtraction formulas for sin θ and cos θ. Multiplication in polar form gives us: z 1 z 2 = r 1 r 2 [ cos(θ1 + θ 2 ) + i sin(θ 1 + θ 2 ) ] (Demonstrated in class. Also explained in the textbook.)

29 Multiplication in polar form gives us: z 1 z 2 = r 1 r 2 [ cos(θ1 + θ 2 ) + i sin(θ 1 + θ 2 ) ] Example: z = 3 + i and w = 3 i. Calculate z w using both rectangular form and polar form.

30 What about division? z 1 = r [ ] 1 cos(θ 1 θ 2 ) + i sin(θ 1 θ 2 ) z 2 r 2 How can we show that this is true? As an exercise, calculate r 1 (cos θ 1 + i sin θ 1 ) r 2 (cos θ 2 + i sin θ 2 ) cos θ 2 i sin θ 2 cos θ 2 i sin θ 2

31 Powers of complex numbers We can generalise the multiplication of complex numbers in polar form to obtain a formula for taking powers of complex numbers. For z = r(cos θ + i sin θ) and n a positive integer, we have De Moivre s Theorem: z n = [r(cos θ + i sin θ)] n = r n (cos nθ + i sin nθ) Example: find ( 1 + 3i ) 4.

32 Next time: n-th roots of complex numbers. More about complex numbers: Download and listen to this podcast to learn more about complex numbers: b00tt6b2