The Multiplicative Inverse of a Complex Number

The multiplicative inverse of a complex number exists such that z∙z^-1 = 1. To find the inverse, let z = a+ib and z^-1 = c+id.

Now, when we multiply z and z^-1, we get:

\begin{matrix} z\times z^{-1} & = & \left ( a+ib \right )\left ( c+id \right )\\ & = & ac+iad+ibc+i^2d\\ & = & \left ( ac-bd \right )+i\left ( ad+bc \right ) \end{matrix}

Knowing the result that z∙z^-1 = 1, we can equate the components:

ac-bd = 1
ad+bc = 0

Solving these simultaneously, we can get c and d in terms of a and b, such that...

c=\frac{a}{a^2+b^2}

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