# Complex Numbers

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Complex Numbers – HMC Calculus Tutorial

The complex numbers are an extension of the real numbers containing all roots of quadratic equations. If we define $i$ to be a solution of the equation $x^{2}= -1$, them the set $\mathbb{C}$ of complex numbers is represented in standard form as $$\left\{ a+bi | a,b \in R\right\}.$$ We often use the variable $z=a+bi$ to represent a complex number. The number $a$ is called the real part of $z$: Re $z$ while $b$ is called the imaginary part of $z$: Im $z$. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

We represent complex numbers graphically by associating $z=a+bi$ with the point $(a,b)$ on the complex plane.

#### Basic Operations

The basic operations on complex numbers are defined as follows: \begin{eqnarray*} (a+bi) + (c+di) & = & (a+c) + (b+d)i \\ (a+bi) – (c+di) & = & (a-c) + (b-d)i \\ (a+bi)(c+di) & = & ac + adi + bci + bdi^2 \\ & = & (ac-bd) + (bc+ad)i \end{eqnarray*} $$\frac{a+bi}{c+di} = \frac{a+bi}{c+di}\cdot\frac{c-di}{c-di} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i$$ In dividing $a+bi$ by $c+di$, we rationalized the denominator using the fact that $(c+di)(c-di) = c^2 -cdi +cdi -d^2i^2 = c^2 + d^2$. The complex numbers $c+di$ and $c-di$ are called complex conjugates. If $z=c+di$, we use $\overline{z}$ to denote $c-di$. Viewing $z=a+bi$ as a vector in the complex plane, it has magnitude $$|z| = \sqrt{a^2+b^2},$$ which we call the modulus or absolute value of $z$.

###### Examples
• $(2+3i)(2-3i)=4-6i+6i-9i^2=4+9=13.$
• $|2+3i| = |2-3i| = \sqrt{4+9}=\sqrt{13}.$

#### Polar Form

For $z=a+bi$, let \begin{eqnarray*} a & = & r\cos\theta \\ b & = & r\sin\theta \end{eqnarray*} from which we can also obtain

$\qquad$\begin{eqnarray*} r & = & \sqrt{a^2+b^2} = |z| \\ \tan\theta & = & \frac{b}{a}. \end{eqnarray*}

If you write $$\theta = \tan^{-1}\frac{y}{x},$$ be careful to choose the value for $\theta$ in the correct quadrant.

Then $$z=r\cos\theta + ir\sin\theta$$ and so, by Euler’s Equation, we obtain the polar form $$z=re^{i\theta}.$$

Euler’s Equation: $$e^{i\theta}=\cos\theta + i\sin\theta$$

Here, $r$ is the magnitude of $z$ and $\theta$ is called the argument of $z$: arg $z$. The argument is not unique; we can add multiples of $2\pi$ to $\theta$ without changing $z$. We define Arg $z$, the principal value of the argument, to be in $(-\pi,\pi]$. The principal value is unique for each $z$ but creates unavoidable, yet interesting, complications due to its discontinuity across the negative real axis where it jumps from $\pi$ to $-\pi$. This jump is called a branch cut.

###### Examples
• $e^{i\pi} = \cos\pi + i\sin\pi = -1$
• $3e^{i\pi/2} = 3(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}) = 3i$
• $2e^{i\pi/6} = 2(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) = \sqrt{3} + i$

Multiplication and division of complex numbers is amazingly simple in polar form! If $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$, then \begin{eqnarray*} z_1z_2 & = & r_1r_2e^{i(\theta_1 + \theta_2)} \\ \frac{z_1}{z_2} & = & \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} \end{eqnarray*} If $z=re^{i\theta}$, then $\overline{z}=re^{-i\theta}$. Do you see why? So $z\overline{z} = (re^{i\theta})(re^{-i\theta}) = r^{2}$.

###### Example

To calculate $(1+i)^{8}$, we can first rewrite $1+i$ as $\sqrt{2}e^{i\pi/4}$. Then \begin{eqnarray*} \left(\sqrt{2}e^{i\pi/4}\right)^{8} & = & (\sqrt{2})^{8}e^{i8\pi/4} \\ & = & 16e^{2\pi i} \\ & = & 16. \end{eqnarray*}

\begin{eqnarray*} \sqrt{1^2+1^2} & = & \sqrt{2} \\ \tan^{-1} \left( \frac{1}{1} \right) & = & \frac{\pi}{4} \end{eqnarray*}

#### Roots of Unity

The equation $$z^{n}=1$$ has $n$ complex-valued solutions, called the $n^{th}$ roots of unity. Since we know each root has magnitude 1, let $z=e^{i\theta}$. Then

\begin{eqnarray*} (e^{i\theta})^{n} & = & 1 \\ e^{in\theta} & = & e^{i(2\pi k)} \\ n\theta & = & 2\pi k \\ \theta & = & \frac{2 \pi k}{n} \end{eqnarray*} so the $n^{th}$ roots of unity are of the form $$z=e^{i\frac{2 \pi k}{n}}.$$

$(e^{i\theta})^{n} = e^{in\theta}$ together with Euler’s Equation, gives us deMoivre’s Formula: $$(\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta$$

$$1=e^{0i}=e^{2\pi ki}$$ for $k=0,\pm 1,\pm2,\ldots$

There are $n$ distinct roots, after which we start dumplicating roots already found.

These are evenly spaced around the unit circle.

###### Example

The $3rd$ roots of unity are \begin{eqnarray*} 1 & & \\ e^{i\frac{2\pi}{3}} & = & -\frac{1}{2} + i\frac{\sqrt{3}}{2} \\ e^{-i\frac{2\pi}{3}} & = & -\frac{1}{2} – i\frac{\sqrt{3}}{2} \end{eqnarray*} You can verify that $(-\frac{1}{2}+ i\frac{\sqrt{3}}{2})^{3} = 1$ and $(-\frac{1}{2} – i\frac{\sqrt{3}}{2})^{3} = 1$.

This tutorial has reviewed the basics of complex arithmetic. The methods of complex analysis, which build on this background, are both intriguing and powerful!

### Key Concept

Standard Form
$\begin{array}{rcl} z & = & a +bi \\ a & = & \mbox{Re } z \\ b & = & \mbox{Im } z \\ |z| & = & \sqrt{a^2+b^2} \\ \overline{z} & = & a-bi \end{array}$

$\begin{array}{rcl} a & = & r\cos\theta \\ b & = & r\sin\theta \\ r & = & \sqrt{a^2 + b^2} \\ \tan\theta & = & \frac{b}{a} \end{array}$

Polar Form
$\begin{array}{rcl} z & = & re^{i\theta} \\ r & = & |z| \\ \theta & = & \mbox{arg } z \\ \overline{z} & = & re^{-i\theta} \end{array}$

Euler’s Equation, $$e^{i\theta} = \cos\theta + i\sin\theta,$$ provides the connection between these two representations of complex numbers.