Complex Numbers

Question 1

How to find complex number in roots of polynomials

Answer 1

Lets say we have x^2 + 1.
Since the degree of this polynomial is 2, there must be 2 roots, however determining these roots can be rather complicated, since no real number squared can be a negative (since we are adding 1, it has to be negative for it to be 0).
We then just declare these roots to be i, and -i, establishing that i^2 = -1, and -i^2 = -1. These are part of an imaginary class of numbers, in which we can say (-)i^2 + 1 = 0.

Another example would be x^2 + 9, in which it would be 3i, and -3i. The way we get 3 is just simply:
If we have x^2 + 9, then that means x^2 must equal -9, since we need the polynomial to equal 0. However, we want x, so we want to find the square root of -9.
We can then split this up into -1 and 9, and take the sqrt of them:
sqrt(-1) = +- i.
sqrt(9) = 3
Therefore it equals 3i, or -3i.

Question 2

Re and Im

Answer 2

Re(z) = the real part of the polynomial.
Im(z) = the imaginary part of the polynomial.

Question 3

Addition

Answer 3

z = u + v, where u, v are polynomials.
Re(z) = Re(u) + Re(v).
Im(z) = Im(u) + Im(v).
Example:
u = 3+4i ; v = 1 - 2i
3 + 1 = 4
4 - 2 = 2
z = 4 + 2i

Question 4

Complex Conjugate

Answer 4

z̄
Re(z̄) = Re(z)
Im(z̄) = -Im(z)
Example:
z = 4 + 2i
Re(z̄) = 4
Im(z) = -2
z = 4 - 2i

Question 5

Modulus

Answer 5

|z|
√Re(z)^2 + Im(z)^2
Example:
z = 3 + 4i
√9 + 16 = √25 = 5

Question 6

Scalar Multiplication

Answer 6

z = a . u
Re(z) = a . Re(u) ; Im(z) = a . Im(u)
Example:
u = 3 + 4i
z = u . 2
3 . 2 = 6
2 . 4i = 8i
z = 6 + 8i

Question 7

Complex Multiplication

Answer 7

z = u . v
Re(z) = Re(u) . Re(v) - Im(u) . Im(v) (the reason for the negative is because of the i^2, since it eventually equals into the negatives due to square rooting principle we mentioned earlier)
Im(z) = Re(u) . Im(v) + Im(u) . Re(v), adding the multiplication of i at the end.

Question 8

Complex Division

Answer 8

z = 1 / u
z = 1 / u = ū / |u|^2
Example:
u = 3 + 4i
ū = 3 - 4i
|u|^2 = 9 + 16 = 25
3 - 4i / 25

Question 9

Way of Representation: Matrix

Answer 9

A way of representing z = a + ib would be:
a -b
b a
in terms of a matrix.

Question 10

Matrix Addition

Answer 10

Mu + v = Mu + Mv

Question 11

Matrix Conjugate

Answer 11

Mz̄ = MTz, the transpose of Mz

Question 12

Matrix Modulus

Answer 12

|z| = detMz, the determinant of Mz

Question 13

Matrix Multiplication

Answer 13

Mu . v = Mu . Mv = Mv . Mu = Mv . u

Question 14

Way of Representation: 2-vector

Answer 14

This is more done by diagram (called the Argand Diagram), with the “y-axis” being represented by the imaginary part, and the “x-axis” being represented by the real part.
We can then show the negative and positive imaginary part via two vectors.

Question 15

2-Vector Advantages (addition, multiplication etc)

Answer 15

We see that the addition of complex numbers mimics vector addition. If we refer back to the page “vectors”, we see the representation of z = x + y, which is what we can input here. We can also see that the conjugate is a reflection in the Re(z) axis, as we again just switch the positive imaginary part with the negative imaginary part. Finally, modulus is just the standard vector size, another easy representation.

Question 16

2-Vector Disadvantages

Answer 16

However, in these diagrams, multiplication and division does not seem to have a natural geometric analogy, as seen by the diagram as there is not really a representation of two vector added together. There are also subtleties which are lost by treating complex numbers as another way of dealing with two vectors. However, some of these issues can be resolved by looking at polar coordinates.

Question 17

Ways of Representation: Polar Coordinates

Answer 17

We have a similar representation to 2-vector representation, however, we also have an angle and modulus to consider.
We used:
z = (r, θ)
with r being the modulus, calculated by |z|, and θ being args z, aka the phase of z, being measured from the real number (x-axis) to the “standard position” of the 2-vector.
Args z is given by:
cos^-1(Re(z) / |z|) = sin^-1(Im(z) / |z|)

Question 18

Ways of Representation: Euler Form

Answer 18

z = r . e^iθ = r(cos θ + i sin θ), also known as Euler’s Formula (e being exponential)
This also leads to:
For all of a that exists in R : (cos θ + i sin θ)^a = cos(aθ) + i . sin(aθ)

Question 19

Euler Form : Multiplication

Answer 19

For example:
Lets say we have u and v, and we want to multiply them.
The phase of u = N, and the phase of v = M
u = se^iN
v = te^iM
u . v = (st)e^i(N+M)

Quick note, the reason why we have (st)e and then i(N + m) is because with s and t, we are multiplying, with iN and iM, we are adding.

Question 20

Euler Form Rules

Answer 20

“The phase resulting from the product of two complex numbers is the result of adding the phase of each” -> simply enough, lets just say that z was the final result, The phase of z would be the addition of the phases of the two other complex number.

“Infinitely many representations” -> this is because sin and cos are called periodic; the values at particular parts repeat when multiplying 2pi.

“The principle value of args z = θ is that for which 0 <= θ < 2π