# 14.1 Introduction

What happens when you square root a negative number? Well, according to a bunch of people in history, nothing. They claimed it was like dividing by zero, non possibilis11 1 Not possible!!

Modern mathematicians, however, think that things do happen when you square root negative numbers (and this can easily be seen after looking at real-world applications).

To denote the square root of a negative number, we declare a new symbol, $i$ 22 2 Note that some people also write this using the letter $j$., which is defined as

$i^{2}=-1$ (14.1)

Using this definition, we can (for example) write

 $\displaystyle\sqrt{-1}=i$ $\displaystyle\sqrt{-15}=\sqrt{15}\sqrt{-1}=\sqrt{15}i$ $\displaystyle\sqrt{-49}=\sqrt{49}\sqrt{-1}=7i$ $\displaystyle 15+\sqrt{-49}=15+\sqrt{49}\sqrt{-1}=7i$ $\displaystyle x+y\sqrt{-1}=x+yi$

In general, we can write a complex number as $x+yi$. Here $x$ denotes the "real" part of the complex number (as there is no $i$ anywhere to be seen) and the $y$ denotes the imaginary part (as there is an $i$).

For a complex number, $z$, we can write the real part of $z$ as $\Re(z)$ and the imaginary part of $z$ as $\Im(z)$. Note that many (most?) people don’t use the weird symbols that this document uses, but instead $\text{Re}(z)$ and $\text{Im}(z)$ to denote the real and imaginary components of a complex number respectively.

Using this definition, we can work out what happens when we do some basic operations.

$(a+bi)+(c+di)=(a+c)+(b+d)i$

We can also multiply them

 $\displaystyle\frac{a+bi}{c+di}$ $\displaystyle=\frac{a+bi}{c+di}\cdot\frac{c-di}{c-di}$ $\displaystyle=\frac{ac-adi+bci-bdi^{2}}{c^{2}-d^{2}i^{2}}$ $\displaystyle=\frac{ac+bd+(bc-ad)i}{c^{2}+d^{2}}$

## 14.1.1 The complex conjugate

The complex conjugate of $x+yi$ is just $x-yi$ - that is, when we take the complex conjugate of a number, the real part remains unchanged, but the sign of the complex part is flipped (i.e. positive numbers become negative and vice versa).

The complex conjugate can be denoted in a number of ways. For a complex number $z$, we can write the conjugate as $\overline{z}$ (it is also sometimes written as $z^{*}$).

###### Theorem 14.1.1

Let $z$ be a complex number, then

$z+\overline{z}=2Re(z)$ (14.2)

This is a useful theorem (well it shows up sometimes), and the proof uses the standard method for proving identities (Section 6.2).

 $\displaystyle z+\overline{z}$ $\displaystyle=x+iy+x-iy$ (14.3) $\displaystyle=2x+iy-iy$ (14.4) $\displaystyle=2x$ (14.5) $\displaystyle=2Re(x)$ (14.6)

## 14.1.2 Dividing complex numbers

Another operation we can perform is division. To divide two complex numbers (which we can call $w$ and $z$), we do something similar to rationalising the denominator (except for complex numbers). Note that $\sqrt{-1}$ behaves just like a surd (except that it’s not in the set of real numbers), so it shouldn’t come as any great surprise that the same techniques are useful here as for surds.

###### Example 14.1.1

Divide $1+i$ by $3+4i$.

We want to make the denominator real, so we just rationalise it,

 $\displaystyle\frac{1+i}{3+4i}$ (14.7) $\displaystyle=\frac{1+i}{3+4i}\cdot\frac{1+i}{3-4i}$ (14.8) $\displaystyle=\frac{(1+i)(3-4i)}{25}$ (14.9)

## 14.1.3 The modulus of a complex number

###### Definition 14.1.1

We define the \saymodulus of a complex number $z=x+iy$ as

$|z|=\sqrt[]{x^{2}+y^{2}}$ (14.10)

If we plot a complex number, then the modulus is the \saylength of the complex number. This is exactly the same as how the magnitude of the vector in $\mathbb{R}^{2}$ (all real-number valued coordinates. e.g. $(1,2)$, $(\pi,6)$, etc.) is defined as

$\left\lvert\begin{pmatrix}x\\ y\end{pmatrix}\right\rvert=\sqrt[]{x^{2}+y^{2}}$ (14.11)
###### Theorem 14.1.2

For all $z\in\mathbb{C}$, we have

$z\overline{z}=\left\lvert z\right\rvert^{2}.$ (14.12)

That is, the value of $z$ multiplied by its complex conjugate is the modulus of $z$.

Proof: we can prove this by direct computation. If $z$ is a complex number then there exist $x$ and $y$ such that $z=x+iy$, and therefore

 $\displaystyle z\overline{z}$ $\displaystyle=(x+iy)\times(x-iy)$ (14.13) $\displaystyle=x^{2}-(iy)^{2}$ (14.14) $\displaystyle=x^{2}+y^{2}$ (14.15)

Note that we used the difference of two squares identity (see Section 6.2) to show this.

###### Theorem 14.1.3

Let $z,w\in\mathbb{C}$, then

 $\displaystyle\overline{zw}=\overline{z}\times\overline{w}$ (14.16)

Proof: let $z=a+bi$ and $w=c+di$, then we proceed with the standard method of proving identities (Section 6.2),

 $\displaystyle\overline{z}\overline{w}$ $\displaystyle=(a-bi)(c-di)$ (14.17) $\displaystyle=ac-(ad+bc)i-bd$ (14.18) $\displaystyle=ac-bd-(ad+bc)i$ (14.19)

What remains is to prove that

$ac-bd-(ad+bc)i=\overline{zw}$ (14.20)

We can work backwards here, and obtain that

 $\displaystyle\overline{zw}$ $\displaystyle=\overline{(a+bi)(c+di)}$ (14.21) $\displaystyle=\overline{ac+(ad+bc)i-bd}$ (14.22) $\displaystyle=\overline{ac-bd+(ad+bc)i}$ (14.23) $\displaystyle=ac-bd-(ad+bc)i.$ (14.24)

Which is what we needed to prove (note that it is of paramount importance that our proof - which it does - flows in both directions; that is, that every step is reversible, see Section 6.2 for more on this).

## 14.1.4 Properties of the complex conjugate and absolute value

• $(zw)^{*}=z^{*}w^{*}$

• $\left(\frac{z}{w}\right)^{*}$

• $\left(z^{*}\right)^{*}$

• $\left\lvert z\right\rvert=\left\lvert z^{*}\right\rvert$

• $\left\lvert zw\right\rvert=\left\lvert z\right\rvert\left\lvert w\right\rvert$

• $\left\lvert\frac{z}{w}\right\rvert=\frac{\left\lvert z\right\rvert}{\left% \lvert w\right\rvert}$

• $\left\lvert z+w\right\rvert\leqq\left\lvert z\right\rvert+\left\lvert w\right\rvert$

## 14.1.5 Some example complex number problems

###### Example 14.1.2

Prove that

$\left\lvert z^{2022}e^{i\theta}\right\rvert=\left\lvert\overline{z}e^{-i\theta% }\right\rvert^{2022}$ (14.25)

where $\overline{z}$ denotes the complex conjugate of $z$.

We can prove this using the standard method for proving identities (see Section 6.2 for more details of the method) - pick one side and work to the other side using reversible steps.

 $\displaystyle|z^{2022}e^{i\theta}|$ $\displaystyle=|z^{2022}||e^{i\theta}|$ (14.26) $\displaystyle=|(z^{*})^{2022}||e^{i\theta}|$ (14.27) $\displaystyle=|(z^{*})^{2022}||1|$ (14.28) $\displaystyle=|(z^{*})^{2022}||1|^{2022}$ (14.29) $\displaystyle=|(z^{*})^{2022}||e^{i\theta}|^{2022}$ (14.30) $\displaystyle=|(z^{*})^{2022}||e^{-i\theta}|^{2022}$ (14.31) $\displaystyle=|(z^{*})|^{2022}|e^{-i\theta}|^{2022}$ (14.32) $\displaystyle=\Big{(}|(z^{*})||e^{-i\theta}|\Big{)}^{2022}$ (14.33) $\displaystyle=|(z^{*})e^{-i\theta}|^{2022}$ (14.34)