Use polar complex numbers to find multiplicative inverse

Use the polar form of complex numbers to show that every complex number $z\neq0$ has multiplicative inverse $z^{-1}$.

If $z=a+bi$, then the polar form is $z=r(cos(\alpha))+i(sin(\alpha))$.

I can do it, not using polar coordinates: Let $z=a+ib$.

Since $1+0i$ is the multiplicative identity, if $x+iy$ is the multiplicative inverse of $z=a+ib$, then $(a+ib)(x+iy)=1+i0$ $\Rightarrow$ $(ax-by)+i(ay+bx)=1+i0$ $\Rightarrow$ $ax-by=1bx+ay=0$ $\Rightarrow$ $x=\frac{a}{a^2+b^2}$, $y=\frac{-b}{a^2+b^2}$, if $a^2+b^2\neq0$.

The multiplicative inverse of $a+ib$ is therefore $\frac{a}{a^2+b^2} +i\frac{-b}{a^2+b^2}$.

Now, should I use this same process and just replace $z=a+ib$ with $z=r(cos(\alpha))+i(sin(\alpha))$?? Or is there another way to go about this process?

• what is cool is that $\frac{1}{r (\cos(\alpha)+i\sin(\alpha))} = \frac{\cos(-\alpha) +i\sin(-\alpha)}{r}$ is true even when $\alpha$ and $r$ are complex numbers – reuns Feb 7 '16 at 0:03

Most of the time complex numbers are introduced using $z=a+i b$ along with the basic addition rule and some more complicated multiplication rules. But really, it is much more natural to think of them in polar coordinates when doing multiplication; especially if you keep them in the form $z=r e^{i \phi}$.
You're right to start with the multiplicative identity $1 \cdot z=z$. If we look at the unit circle, $1=1+i(0)$, and look what is happening when we multiply by $z$ we see two things: the radius changes from $1\rightarrow r$, and the angle rotates from $0\rightarrow \phi$. Therefore, to find inverse we need to find some $z'$ such that $1 \cdot z \cdot z'=1$. This $z'$ operation will also be a scaling and a rotation operation, but in this case, what angle should you rotate by, and what value should you scale by, to return to 1?
Once this concept clicks, you can really see that $z=r e^{i \phi}$ is the natural choice for complex multiplication, and makes it almost as easy as $z=a+ib$ for complex addition.
Yes, you can directly check that inverse of $r(\cos \alpha +i\sin \alpha)$ is $\frac{1}{r}(\cos \alpha - i \sin \alpha)$ (try to multiply them and use $\sin^2 \alpha+\cos^2 \alpha=1$).