---
author: Alvie Rahman
date: \today
title: MMME1026 // Complex Numbers
tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, complex_numbers ]
---

# Complex Numbers

## What is a Complex Number?

- $i$ is the unit imaginary number, which is defined by:

  $$ i^2 = -1 $$

- An arbritary complex number is written in the form

  $$z = x + iy$$

  Where:

  - $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
  - $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)

- Two complex numbers are equal if both their real and imaginary parts are equal

  e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$

### The Complex Conjugate

Given complex number $z$:

$$z = z + iy$$

The complex conjugate of z, $\bar z$ is:

$$\bar{z} = z -iy$$

### Division of Complex Numbers

- Multiply numerator and denominator by the conjugate of the denominator

<details>
<summary>

#### Example

</summary>

> \begin{align*}
  z_1 &= 5 + i \\
  z_2 &= 1 -i \\
  \\
  \frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
    &= \frac{5 + i + 5i -1}{1 + 1} \\
    &= \frac{4 + 6i}{2} = 2 + 3i
> \end{align*}

</details>

### Algebra and Conjugation

When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after
doing the algebraic operations:

\begin{align*}
\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\
\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\
\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}}
\end{align*}

The conjugate of a real number is the same as that number.

#### Application

If $z$ is a root of the polynomial equation

$$0 = az^2 + bz + c$$

with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because

\begin{align*}
0 &= \overline{az^2 + bz + c} \\
  &= \bar{a}\bar{z}^2 + \bar{b}\bar{z} + \bar{c} \\
  &= a\bar{z}^2 + b\bar{z} + c
\end{align*}

### The Argand Diagram

A general complex number $z = x + iy$ has two components so it can can be represented as a point in
the plane with Cartesion coordinates $(x, y)$.

\begin{align*}
4-2i &\leftrightarrow (4, -2) \\
-i &\leftrightarrow (0, -1) \\
z &\leftrightarrow (x, y) \\
\bar z &\leftrightarrow (x, -y)
\end{align*}

### Plotting on a Polar Graph

We can also describe points in the complex plain with polar coordinates $(r, \theta)$:

\begin{align*}
z &= r(\cos\theta + i\sin\theta) &\text{ polar form of $z$} \\
r &= \sqrt{x^2+y^2} &\text{(modulus)}\\
\theta &= \arg z,\text{ where} \tan \theta = \frac y x &\text{(argument)} \\
x &= r\cos \theta \\
y &= r\sin \theta
\end{align*}

Be careful when turning $(x, y)$ into $(r, \theta)$ form as $\tan^{-1} \frac y x = \theta$ does not
always hold true as there are many solutions.

#### Choosing $\theta$ Correctly

1. Determine which quadrant the point is in (draw a picture).
2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
   If it puts you in the wrong quadrant, add or subtract $\pi$.

## Exponential Functions

- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:

  $$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$

- The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
- Note the important properties:

\begin{align*}
e^{a+b} &= e^a e^b \\
(e^a)^b &= e^{ab} 
\end{align*}

## Euler's Formula

$$e^{i\theta} = \cos\theta + i\sin\theta$$

- Properties of $e^{i\theta}$: For any real angle $\theta$ we have

  $$|e^{i\theta}| = |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$

  and 

  $$ \arg {e^{i\theta}} = \theta $$

- A complex number in *polar form*, where $r = |z|$, and $\theta = \arg z$, may alternatively be
  written in its *exponential form*:

  $$z = re^{i\theta}$$

  **Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$

<details>
<summary>

### Example 1

Write $z = -1 + i$ in exponential form

</summary>

> $\arg z = \frac {3\pi} 4$
> $|z| = \sqrt 2$
> 
> So $z = \sqrt2e^{i\frac{3\pi} 4}$

</details>

<details>
<summary>

### Example 2

The equations for a mechanical vibration problem are found to have the following mathematical
solution:

$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$

</summary>

where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
constants.
Although $z(t)$
is complex and cannot directly represent a physical solution, it turns out that the real and
imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
this physical information efficiently as follows:

a. Put the denominator in the form

   $$ae^{i\delta}$$

   where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
   and $\gamma$.


   > \begin{align*}
     a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
     \delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
   > \end{align*}

b. Hence find the constants $b$ and $\varphi$ such that

   $$x(t) = b\cos(\omega t + \varphi)$$

   and write a similar expression for $y(t)$.

   > \begin{align*}
     z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
     x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
     \therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
     \Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
     \\
     b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
     \varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
     \\
     y(t) &= \frac 1 a \sin(\omega t - \delta) \\
   > \end{align*}

</details>

## Products of Complex Numbers

Suppose we have 2 complex numbers:

$$z_1 = x_1 + iy_1 = r_1e^{i\theta_1}$$  
$$z_2 = x_2 + iy_2 = r_2e^{i\theta_2}$$

Using $e^a e^b = e^{a+b}$, the product is:

\begin{align*}
z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
&= r_1r_2e^{i\theta_1}e^{i\theta_2} \\
&= r_1r_2e^{i(\theta_1+\theta_2)} \\
\\
|z_1z_2| &= |z_1|\times|z_2| \\
\arg z_1z_2 &= \arg z_1 \times \arg z_2
\end{align*}

## de Moivre's Theorem

Let $z = re^{i\theta}$. Consider $z^n$.

Since $z = r(\cos\theta + i\sin\theta)$,
\begin{align*}
z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
\end{align*}

But also

\begin{align*}
z^n &= (re^{i\theta})^n \\
    &= r^n(e^{i\theta})^n \\
    &= r^ne^{in\theta} \\
    &= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
\end{align*}

By equating (1) and (2), we find de Moivre's theorem:

\begin{align*}
r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
\end{align*}

<details>
<summary>

### Example 1

Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.

</summary>

> \begin{align*}
  r &= |1+i| = \sqrt2 \\
  \theta &= \arg{1+i} = \frac \pi 4 \\
  \\
  \text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
  (i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
             &= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
             &= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
             &= 2^7 (1 - i) \\
             &= 128 - 128i
> \end{align*}
</details>

<details>
<summary>

### Example 2

Use de Moivre's theorem to show that

\begin{align*}
\cos{2\theta} &= \cos^2\theta-\sin^2\theta \\
\text{and} \\
\sin{2\theta} &= 2\sin\theta\cos\theta
\end{align*}

</summary>

> Let $n=2$:

> \begin{align*}
  (\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
  \text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
  \text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
> \end{align*}

</details>

<details>
<summary>

### Example 3

Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.

</summary>

> \begin{align*}
  r &= \sqrt{2} \\
  \arg \omega = \theta &= \frac 3 4 \pi \\
  \\
  \omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
  \bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
  \omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
                          &= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
> \end{align*}

</details>

## Complex Roots of Polynomials

<details>
<summary>

### Example

Find which complex numbers $z$ satisfy

$$z^3 = 8i$$

</summary>

> 1. Write $8i$ in exponential form,
> 
>    $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
> 
>    $\therefore 8i = 8e^{i\frac \pi 2}$
> 
> 
> 2. Let the solution be $r = re^{i\theta}$.
> 
>    Then $z^3 = r^3e^{3i\theta}$.
> 
> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
> 
>    i. Compare modulus:
> 
>       $r^3 = 8 \rightarrow r = 2$
> 
>    ii. Compare argument:
> 
>        $$3\theta = \frac \pi 2$$
> 
>        is a solution but there are others since 
> 
>        $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
> 
>        so we get a solution whenever
> 
>        $$3\theta = \frac \pi 2 + 2n\pi$$
> 
>        for any integer `n`
> 
>        - $n = 0 \rightarrow z = \sqrt3 + i$
>        - $n = 1 \rightarrow z = -\sqrt3 + i$
>        - $n = 2 \rightarrow z = -2i$
>        - $n = 3 \rightarrow z = \sqrt3 + i$
>        - $n = 4 \rightarrow z = -\sqrt3 + i$
>        - The solutions start repeating as you can see
> 
>        In general, an $n$-th order polynomial has exactly $n$ complex roots.
>        Some of these complex roots may be real numbers.
> 
> 4. There are three solutions

</details>
