complex numbers - Why: $z^{-1}=cos (-phi)+isin(-phi)$

Let's suppose that we have a complex number with

$$r=1 \implies z=\cos\phi+i\sin\phi$$

Then why is $$z^{-1} =\cos (-\phi)+i\sin (-\phi)=\cos\phi-i\sin\phi$$

Answer

This is De Moivre's formula
in action:
For complex $z$ and any integer $n$ we have
$$\begin{align*} z^n=(\cos (\varphi)+i\sin(\varphi))^n=\cos(n\varphi)+i\sin(n\varphi) \end{align*}$$

The formula
$$\begin{align*} z^{-1}=\cos(-\varphi)+i\sin(-\varphi) \end{align*}$$
is De Moivre's formula evaluated at $n=-1$.