Analysis Complex numbers

Question 1

Complex numbers

Answer 1

A complex number is an expression of the form z = x + y . i where x,y ∈ R and is is the imaginary unit. Important: i^2 = -1
We call x the real part of z and denote Re(z)
We call y the imaginary part of z and denote it Im(z)

Question 2

Modules and argument

Answer 2

A complex number can also be defined by its length and angle. Suppose z = x + y . i Then z can be rewritten as z = r(cos(phi) + isin(phi)). r is called the modulus of z. r describes the distance from origin (0,0) to the point (x,y) i.e. the length. Theta is an argument of z which is the angle with the x-axis. If theta ∈ (-pi, pi] so -pi< theta<pi then theta is the principal argument of z denoted by Arg(z)

Question 3

complex conjugate

Answer 3

Every complex number z has its complex conjugate( z bar)
Suppose z = x + y.i.
Then zbar is defined by: zbar = x - y.i
Note that |zbar| = |z|, arg(zbar) = -Arg(z), Re(z) = Re(zbar) and Im(zbar) = -Im(z)

Question 4

Moivre’s theorem

Answer 4

(cos(theta) +isin(theta))^n = (cos(ntheta) + isin(ntheta))

When we use it on complex numbers
z^n = (x + y . i)^n = (r(cos(theta) + isin(theta)))^n =
r^n(cos(ntheta) + isin(ntheta))

Question 5

Solving higher order equations with complex solutions

Answer 5

When we want to solve z^k = w, with w a complex number, do the following
1. write w with polar coordinates, so you get z^k = r(cos(theta) + isin(theta))
2. The modili of all solutions are the same: r^(1/k)
3. We can write the argument of solutions j as theta/k + (j -1) . 2pi/k
4. In the end, solution j can be written as zj = r^(1/k)(theta/k + (j -1) * (2pi/k), j ∈ {1,……,k}. You always have k solutions, equal to the degree of the order