My book states that:
The inverse of a complex number $z = \rho \cdot cis \theta$, not zero, is a complex number $\color{red}{\text{$\frac{1}{z}= \frac{1}{\rho}\cdot cis(-\theta)$}}$.
I don't understand the red part. I tried:
$$\frac{1}{\rho \cdot cis(\theta)} = \frac{1}{\rho \cdot (\cos \theta+i\sin\theta)} = \frac{1}{\rho}\cdot\frac{1}{\cos \theta+i\sin\theta} = ???$$
What happens next?
The identity $e^{ix}=\cos(x)+i\sin(x)=\text{cis}(x)$ makes this clearer. (This can be proven with taylor series). Then if $z=\rho e^{ix}$, we have $z^{-1}=\rho^{-1}e^{-ix}=\rho^{-1}\text{cis}(-x)$.
To see this directly though, we have $$\frac{1}{\rho}\frac{1}{\cos(x)+i\sin(x)}=\frac{1}{\rho}\frac{1}{\cos(x)+i\sin(x)}\frac{\cos(x)-i\sin(x)}{\cos(x)-i\sin(x)}=\frac{1}{\rho}\frac{\cos(x)-i\sin(x)}{\cos^2(x)+\sin^2(x)}=\frac{\cos(x)-i\sin(x)}{\rho}$$
We have $$ \frac 1{\operatorname{cis}(\theta)} = \frac{1}{\cos \theta + i \sin \theta} = \frac{1}{\cos \theta + i \sin \theta} \cdot \frac{\cos \theta - i \sin \theta}{\cos \theta - i \sin \theta}\\ = \frac{\cos \theta - i \sin \theta}{\cos^2 \theta - i^2 \sin^2 \theta} = \frac{\cos \theta - i \sin \theta}{\cos^2 \theta + \sin^2 \theta} \\ = \frac{\cos \theta - i \sin \theta}{1} = \operatorname{cis}(-\theta) $$
