Proof of multiplicative inverse for polar complex numbers

Question

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: If $z=a+ib$.

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 $a^2+b^2\neq0$.

Thus, the multiplicative inverse of $a+ib$ is $\frac{a}{a^2+b^2} +\frac{-b}{a^2+b^2}$.

Now, how can I use this information to find it using $z=r(\cos(\alpha)+i\sin(\alpha))$??

I'm stuck...

Answer 1

If you write any non-zero complex number as $z = r e^{i \theta}$ where $r$ and $\theta$ are reals and $r \ne 0$, then $w = \frac1{r} e^{-i \theta}$ satisfies $z w = 1$.

You can write this in terms of $\sin$ and $\cos$, but I prefer this if you know that $e^{i a} = \cos(a) + i \sin(a) $.

Answer 2

It is basically the same calculation, $\frac{1}{r}\left(\cos\theta-i\sin\theta\right)$.

We can also view things geometrically. Multiplying by $r\left(\cos\theta+i\sin\theta\right)$ is rotating counterclockwise by $\theta$, and scaling by a factor $r$. To "undo" this, we rotate clockwise by $\theta$, or equivalently counterclockwise by $-\theta$, and scale by the factor $\frac{1}{r}$. That gives the formula of the first sentence.