MAT01A1: Complex Numbers (Appendix H)

Transcription

1 MAT01A1: Complex Numbers (Appendix H) Dr Craig 14 February 2018

2 Announcements: e-quiz 1 is live. Deadline is Wed 21 Feb at 23h59. e-quiz 2 (App. A, D, E, H) opens tonight at 19h00. Deadline is Thu 22 Feb at 23h59. Saturday class this week: 09h00 to 12h00 in D1 LAB 110. Please come along if you would like additional assistance with any of the topics that we have covered so far.

3 Lecturers Consultation Hours Monday: 12h00 13h30 Ms Richardson (C-503) Wednesday: 15h00 16h00 Ms Richardson (C-503) Thursday: 11h20 12h55 Dr Craig (C-508) Friday: 11h20 12h55 Dr Craig (C-508)

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

5 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.)

6 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

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

8 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

9 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.

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

11 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)

12 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

13 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

14 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

15 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

16 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.

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

18 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.

19 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.

20 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 n.2π.θ = arg(z) where n is any integer (n Z).

21 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].

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

23 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.)

24 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.

25 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

26 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.

27 From powers to roots in C Suppose we want to find the n-th root of z = r(cos θ + i sin θ). That is, we want the complex number w = s(cos ϕ + i sin ϕ) such that w n = z. From De Moivre s Theorem we want s n (cos nϕ + i sin nϕ) = r(cos θ + i sin θ) To get this, we need s n = r and cos nϕ = cos θ and sin nϕ = sin θ Thus nϕ = θ + 2kπ.

28 Roots of a complex number: n-th roots Let z = r(cos θ + i sin θ) and let n be any positive integer. Then z has n distinct n-th roots. That is, for k = 0, 1, 2,..., n 1 the roots are [ ( ) ( ) ] θ + 2kπ θ + 2kπ w k = r 1/n cos + i sin n n All of the roots of z lie on the circle of radius r 1/n in the complex plane.

29 It often helps to think of the argument of a complex root in the following way: θ + 2kπ = θ ( ) 2π n n + k n The argument of the first root will simply be θ n (because k = 0). Each root after that has the same modulus (r 1/n ) but is rotated anti-clockwise by 2π n. You only need to calculate solutions up to k = n 1. If you let k = n then you will have the same complex number as w 0 but with a different argument.

30 Example of roots of a complex number Find the cube roots of z = i. Note that a = 0 and b = 1. Thus we get r = 1 and arg(z) = θ = π 2

31 Solutions to 3 i w 0 = 3 ( ( π ) ( π )) 1 cos + i sin 6 6 w 1 = 3 ( ( π 1 cos 6 + 2π ) ( π + i sin π 3 w 2 = 3 ( ( π 1 cos 6 + 4π ) ( π + i sin π 3 )) )) Exercise: simplify the angles in w 1 and w 2 and convert each solution to the form a + bi. You will see that w 2 is a solution that you might have found by inspection.

32 Another exercise: Take w 0, w 1 and w 2 and cube each of them using the rectangular form (a + bi). Check that in each case you get i as the solution. Also, look at Example 7 on page A62 of the textbook.