Complex Numbers

Introduction

An imaginary is defined as $i$ with the following property

i^{2} = - 1

It corresponds to a $\frac{π}{2} r a d$ anti-clockwise rotation in the Argand plane

A complex number is defined as.

z = a + i b

Euler's Identity

e^{i θ} = \cos θ + i \sin θ

Proof using Maclaurin Series

Proof for Euler's Identity using Maclaurin Series

Alternative Proof using Calculus

Proof for Euler's Identity using ODEs

Polar Form

A vector has an angle $θ$ and a magnitude $r$ . It should be self-evident that we can rewrite this using Eulers Identity.

z = r (\cos θ + i \sin θ) = r e^{i θ}

The angle $θ$ can be sufficiently described using this range.

- π \leq θ \leq π

The magnitude of a complex number can be expressed as $r = \sqrt{a^{2} + b^{2}} = | | z | |$

Conjugate

The conjugate of $z$ is

z^{*} = a - i b

z^{*} = r e^{- i θ}

Thus

| | z | |^{2} = z \cdot z^{*}

Scaling & Rotations

A complex number encodes a rotation and a scaling. When $a$ is multiplied with $b$ where $a, b \in C$

a \cdot b = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}

$a$ acts on $b$ by scaling it by $r_{1}$ and rotating it by $θ_{1}$

Fun Fact

Complex numbers can also help us encode rotations in 3d space! It's called Quaternions, but you'll need 4 dimension, 1 real and 3 imaginary part. It powers most gaming/3d simulation software.

Argument

Argument is basically the angle between the vector and the x-axis.

a r g (z) = a r g (r e^{i θ}) = θ

Hopefully it is obvious that

a r g (z_{1} \cdot z_{2}) = a r g (z_{1}) + a r g (z_{2})

De Moivre's Theorem

Based on the above, it's pretty evident that

z^{n} = r^{n} e^{i n θ} = r^{n} (\cos n θ + i \sin n θ)

Correspondingly

z^{\frac{1}{n}} = r^{\frac{1}{n}} (\cos \frac{1}{n} θ + i \sin \frac{1}{n} θ)

Solving for nth roots

Solve for

z^{n} = C

Fundamental theorem of algebra state that there exist $n$ number of solutions

Express C in polar form
Apply the De Moivre's theorem to solve for $z$
Find the suitable values of $k$

Example

Solve for

z^{5} = 1 + i

Express $1 + i$ in polar form

1 + i = \sqrt{2} \cdot e^{i (\frac{π}{4} + 2 π k)}

Apply De Moivre's theorem

z = (1 + i)^{\frac{1}{5}} = 2^{\frac{1}{10}} e^{i (\frac{π}{20} + \frac{2 k π}{5})}

Find the values of $k$
Since $- π \leq θ \leq π$ . Thus $k \in {- 2, - 1, 0, 1, 2}$
$∴ Q E D$

Expressing $\cos n θ$ , $\sin n θ$ in terms of $\cos^{k} θ$ and $\sin^{k} θ$ .

Expanding the expression using De Moivre theorem and algebra will yield different results. This allows you to equate and express $\cos n θ$ and $\sin n θ$ in unique ways

(\cos θ + i \sin θ)^{n}

Expand it algebraically
Expand it using De Moivre theorem
Equate the reals and imaginaries

Expand $(\cos θ + i \sin θ)^{3}$ . Express $\cos 3 θ$ and $\sin 3 θ$ in terms of $\cos^{k} θ$ and $\sin^{k} θ$ .

Algebraically

\begin{aligned} (\cos θ + i \sin θ)^{3} & = \cos^{3} θ + 3 \cos^{2} θ (i \sin θ) + 3 \cos θ (i \sin θ)^{2} + (i \sin θ)^{3} \\ = (\cos^{3} θ - 3 \cos θ \sin^{2} θ) + i (3 \cos^{2} θ \sin θ - \sin^{3} θ) \end{aligned}

De Moivre theorem

(\cos θ + i \sin θ)^{3} = \cos 3 θ + i \sin 3 θ

Equate the imaginary and real parts.
Equating the real part

\begin{aligned} \cos 3 θ & = \cos^{3} θ - 3 \cos θ \sin^{2} θ \\ = \cos^{3} θ - 3 \cos θ (1 - \cos^{2} θ) \\ = 4 \cos^{3} θ - 3 \cos θ \end{aligned}

Equating the imaginary part:

\begin{aligned} \sin 3 θ & = 3 \cos^{2} θ \sin θ - \sin^{3} θ \\ = 3 (1 - \sin^{2} θ) \sin θ - \sin^{3} θ \\ = - 4 \sin^{3} θ + 3 \sin θ \end{aligned}

Interesting question

Given $n$ lighthouses along a unit circle, suppose an observer is standing inside one of the lighthouse. Find and Express the product of the distance between the observer and all other lighthouses, ( $d_{0} \cdot d_{1} \cdot d_{2} \dots$ ) in terms of $n$ .

Solution

Consider an Argand diagram. The $n$ roots of $x^{n} = 1$ are spaced equidistance apart along a unit circle. Recall roots of unity.

x^{n} = 1

Suppose the observer stands at point $O$ (on the roots) where $O = L_{0} = 1 + 0 i$ .

The $n$ roots $L_{0}, L_{1}, \dots L_{n - 1}$

x^{n} - 1 = (x - L_{0}) (x - L_{1}) (x - L_{2}) \dots (x - L_{n - 1})

Like vectors, see that $x - L_{a}$ gives a complex number whose magnitude is equal to the distance between $x$ and $L_{a}$ . Thus the product

\begin{aligned} d_{0} \cdot d_{1} \cdot d_{2} \dots d_{n - 2} & = | (O - L_{1}) (O - L_{2}) \dots (O - L_{n - 1}) | \\ = \frac{O^{n} - 1}{O - 1} \\ = 1 + O + O^{2} + O^{3} \dots O^{n - 1} \\ = n \end{aligned}

Source: https://www.youtube.com/watch?v=8GPy_UMV-08

Relationship with Matrices

An anti-clockwise rotation of a vector by $θ$ can be described as

R_{θ} = (\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix})

Since $i$ represents a $\frac{π}{2}$ rotation, it can also be expressed as a matrix

i = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})

such that

i^{2} = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) = (\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}) = - 1

A complex number can be rewritten in matrix form as

z = a + i b = (\begin{matrix} a & - b \\ b & a \end{matrix})

In which the conjugate equates to the transpose of the matrix

z^{*} = z^{T}

Split Complex numbers

Are defined as

\sqrt{j} = - 1

j^{2} = 1, j \neq 1 j \neq - 1

Properties

e^{j x} = \cosh x + j \sinh x

Forms a hyperbola
multiplying by $j$ is a reflection about the $y = x$ axis.
and

| j | = i

Introduction

Euler's Identity

Proof using Maclaurin Series

Alternative Proof using Calculus

Polar Form

Conjugate

Scaling & Rotations

Argument

De Moivre's Theorem

Solving for nth roots

Expressing cos⁡nθ, sin⁡nθ in terms of cosk⁡θ and sink⁡θ.

Interesting question

Solution

Relationship with Matrices

Split Complex numbers

Expressing $\cos n θ$ , $\sin n θ$ in terms of $\cos^{k} θ$ and $\sin^{k} θ$ .