7. Identities, inverses and division of complex numbers

• [4.]The other thing that we need to think about is the multiplicative inverse.

  Let's first think about the multiplicative inverse for real numbers. What is the multiplicative inverse of 2? We call it $2^{-1}$ and it's the number that you have to multiply 2 by to get to 1, which is clearly 0.5. We can write this as:

  $$2^{-1}2=1\implies 2^{-1}=0.5$$

  How about for complex numbers? What is the multiplicative inverse of $z=a+bi$, $a,b\in\mathbb{R}$? Let's call it $z^{-1}=c+di$, $c,d\in\mathbb{R}$ and it must have the property such that:

  $$z z^{-1}=z^{-1}z=1$$

  In fact we only need to deal with one of $z z^{-1}$ or $z^{-1}z$ because we saw that complex numbers commute under multiplication.

  Let's figure out what the $c$ and $d$ have to be in terms of $a$ and $b$. Remember, we are given $z$ and we have to figure out what is $z^{-1}$. Let's write the whole thing out:

  $$zz^{-1}=(a+bi)(c+di)=ac-bd+i(ad+bc)$$

  and we want this to equal 1. Well, for this to be 1, we must have that $ac-bd=1$ and $ad+bc=0$. We can solve this for $c$ and $d$ and we find:

  $$c=\frac{a}{a^2+b^2}\, ,\,\,\,\, d=\frac{-b}{a^2+b^2}$$

  or

  $$z^{-1}=\frac{a-bi}{a^2+b^2}$$

  In fact we can therefore write:

  $$z^{-1}=\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}$$

  Clearly, $z^{-1}$ only exists if $a$ and $b$ are not both zero, in other words, if $z\neq0$.

You might ask why we can't just write that

$$z^{-1}=\frac{1}{a+bi}$$

Indeed this is quite correct, but it gives you no idea about where this number is in the complex plane. In order to put it into Cartesian Form, you need to know what is the real part, and what is the imaginary part, and that is written then explicitely as:

$$z^{-1}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i$$

or just

$$z^{-1}=\frac{a-bi}{a^2+b^2}$$

This way the number which is the multiplicative inverse of $z=a+bi$ is $\frac{a}{a^2+b^2}$ along the real axis and $\frac{-b}{a^2+b^2}$ along the imaginary axis.

Actually, we can do something kind of strange here. We can think about our first function of complex numbers. We know how to draw the graph of $y=1/x$. We do this by setting up the $x$ and $y$ axes, and finding the points in the $(x,y)$ plane such that $y=\frac{1}{x}$. What would this mean for complex numbers. Well for real numbers, each axis was one dimensional, and so we ended up with a two dimensional graph. Here, we have a mapping from points in a two dimensional space to points in a two dimensional space, ie:

$$f:a+bi\rightarrow \frac{a-bi}{a^2+b^2}$$

so it kinda looks like this should be four dimensional, ie. all points $(z,w)$ such that $w=\frac{1}{z}$

Well, we can't do this, so instead we put two complex planes next to each other, and ask where entire lines of complex numbers get mapped to. The easiest lines to think about are lines of constant imaginary part (ie. horizontal lines) and lines of constant real part (ie. vertical lines). It turns out that these lines of points get mapped in a really pretty way to circles of different positions and radii under the map $z\rightarrow \frac{1}{z}$.

There is a mathemafrica blog-post here on plotting functions of complex numbers, the material is not examinable. We recommend reading this blog-post here to see some animated mapping. Below are some plots generated using Mathematica which you will also find in the blog-post.

In the figure above, the colors of each line are matched, and you can see four points which are mapped to their correspondingly colored points in the complex plane.