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, ii 22 2 Note that some people also write this using the letter jj., which is defined as

i2=1i^{2}=-1 (14.1)

Using this definition, we can (for example) write

15+49=15+491=7i\displaystyle 15+\sqrt{-49}=15+\sqrt{49}\sqrt{-1}=7i
x+y1=x+yi\displaystyle x+y\sqrt{-1}=x+yi

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

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

We can add complex numbers:


We can also multiply them

a+bic+di\displaystyle\frac{a+bi}{c+di} =a+bic+dicdicdi\displaystyle=\frac{a+bi}{c+di}\cdot\frac{c-di}{c-di}

14.1.1 The complex conjugate

The complex conjugate of x+yix+yi is just xyix-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 zz, we can write the conjugate as z¯\overline{z} (it is also sometimes written as z*z^{*}).

Theorem 14.1.1

Let zz be a complex number, then

z+z¯=2Re(z)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).

z+z¯\displaystyle z+\overline{z} =x+iy+xiy\displaystyle=x+iy+x-iy (14.3)
=2x+iyiy\displaystyle=2x+iy-iy (14.4)
=2x\displaystyle=2x (14.5)
=2Re(x)\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 ww and zz), we do something similar to rationalising the denominator (except for complex numbers). Note that 1\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+i1+i by 3+4i3+4i.

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

1+i3+4i\displaystyle\frac{1+i}{3+4i} (14.7)
=1+i3+4i1+i34i\displaystyle=\frac{1+i}{3+4i}\cdot\frac{1+i}{3-4i} (14.8)
=(1+i)(34i)25\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+iyz=x+iy as

|z|=x2+y2|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 2\mathbb{R}^{2} (all real-number valued coordinates. e.g. (1,2)(1,2), (π,6)(\pi,6), etc.) is defined as

|(xy)|=x2+y2\left\lvert\begin{pmatrix}x\\ y\end{pmatrix}\right\rvert=\sqrt[]{x^{2}+y^{2}} (14.11)
Theorem 14.1.2

For all zz\in\mathbb{C}, we have

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

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

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

zz¯\displaystyle z\overline{z} =(x+iy)×(xiy)\displaystyle=(x+iy)\times(x-iy) (14.13)
=x2(iy)2\displaystyle=x^{2}-(iy)^{2} (14.14)
=x2+y2\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,wz,w\in\mathbb{C}, then

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

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

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

What remains is to prove that

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

We can work backwards here, and obtain that

zw¯\displaystyle\overline{zw} =(a+bi)(c+di)¯\displaystyle=\overline{(a+bi)(c+di)} (14.21)
=ac+(ad+bc)ibd¯\displaystyle=\overline{ac+(ad+bc)i-bd} (14.22)
=acbd+(ad+bc)i¯\displaystyle=\overline{ac-bd+(ad+bc)i} (14.23)
=acbd(ad+bc)i.\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*(zw)^{*}=z^{*}w^{*}

  • (zw)*\left(\frac{z}{w}\right)^{*}

  • (z*)*\left(z^{*}\right)^{*}

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

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

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

  • |z+w||z|+|w|\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

|z2022eiθ|=|z¯eiθ|2022\left\lvert z^{2022}e^{i\theta}\right\rvert=\left\lvert\overline{z}e^{-i\theta% }\right\rvert^{2022} (14.25)

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

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.

|z2022eiθ|\displaystyle|z^{2022}e^{i\theta}| =|z2022||eiθ|\displaystyle=|z^{2022}||e^{i\theta}| (14.26)
=|(z*)2022||eiθ|\displaystyle=|(z^{*})^{2022}||e^{i\theta}| (14.27)
=|(z*)2022||1|\displaystyle=|(z^{*})^{2022}||1| (14.28)
=|(z*)2022||1|2022\displaystyle=|(z^{*})^{2022}||1|^{2022} (14.29)
=|(z*)2022||eiθ|2022\displaystyle=|(z^{*})^{2022}||e^{i\theta}|^{2022} (14.30)
=|(z*)2022||eiθ|2022\displaystyle=|(z^{*})^{2022}||e^{-i\theta}|^{2022} (14.31)
=|(z*)|2022|eiθ|2022\displaystyle=|(z^{*})|^{2022}|e^{-i\theta}|^{2022} (14.32)
=(|(z*)||eiθ|)2022\displaystyle=\Big{(}|(z^{*})||e^{-i\theta}|\Big{)}^{2022} (14.33)
=|(z*)eiθ|2022\displaystyle=|(z^{*})e^{-i\theta}|^{2022} (14.34)