Why: $z^{-1}=\cos (-\phi)+i\sin(-\phi)$ Let's suppose that we have a complex number with
$$r=1 \implies z=\cos\phi+i\sin\phi$$
Then why is $$z^{-1} =\cos (-\phi)+i\sin (-\phi)=\cos\phi-i\sin\phi$$

Note that sin (-φ)  =  -sin (φ)  and cos (-φ)  = cos (φ)
Thus  $$(\cos (-φ)+i\sin (-φ))(\cosφ-i\sinφ)=$$
$$(cos (-φ)\cosφ -\sin (-φ)\sinφ )+ i(sin (-φ)cosφ-cos (-φ)sinφ)=$$
$$(cos^2φ + sin^2φ) + 0i =1$$
