How to prove $\frac{1}z=\operatorname{cis}(-\theta)$ given $z=\operatorname{cis}(\theta)$

Question

I know $\operatorname{cis}(-\theta)$ equals $\cos(\theta)-i\sin(\theta)$
and $\frac{1}z=\frac{1}\cos+i\frac{1}\sin=\sec+\frac{\csc}i$
(please, correct me if I'm wrong.)

I just don't know how to relate these two now. Can anyone provide some guidance?

Answer 1

It is not correct in general to say if $z=a+bi$ then $\frac 1 z = \frac 1 a + \frac 1 b i; $

rather, $\frac 1 z = \frac a {a^2+b^2} - \frac b {a^2+b^2}i. $

Note that $\cos^2\theta + \sin^2\theta = 1$.

Therefore, if $z=e^{i\theta}=\cos \theta + i \sin \theta$ then $\frac 1 z =e^{-i\theta}=\cos\theta-i\sin\theta.$