MAT01A1: Complex Numbers (Appendix H) Dr Craig 13 February 2019
Introduction Who: Dr Craig What: Lecturer & course coordinator for MAT01A1 Where: C-Ring 508 acraig@uj.ac.za Web: http://andrewcraigmaths.wordpress.com
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 email: check this email account twice per week or set up forwarding to an address that you check frequently.
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
Other announcements No tuts for MAT01A1 on Wednesdays. If you see this on your timetable, it is an error. (To move your Chem. prac., email Mr Kgatshe ckgatshe@uj.ac.za) CSC02A2 students. Email Dr Craig regarding tutorial clash. Maths Learning Centre in C-Ring 512: 10h30 15h25 Mondays 08h00 15h25 Tuesday to Thursday 08h00 12h05 Fridays
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)
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.
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.
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).
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.)
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
More numbers in the complex plane 4 + 2i Im 2 + 3i R 2 2i 3 2i
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
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.
Examples Add: z = 2 7i and w = 4 + 2i. Calculate: (1 + i) (3 4i). Multiply: z = 2 + 3i and w = 4 2i.
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)
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
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
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
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
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 1 +... + 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.
Complex roots of polynomials Every polynomial of degree n has n complex roots. (Roots might be repeated, e.g. x 2 = 0.)
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.
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.
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).
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].
Examples: z = 1 + 3i and w = 1 i. Find z, arg(z), w, arg(w). ( 1, 3) Im r R (1, 1)
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.)
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.
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
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.
Next time: n-th roots of complex numbers. More about complex numbers: Download and listen to this podcast to learn more about complex numbers: https://www.bbc.co.uk/programmes/ b00tt6b2