Question

If z $\ne 0,$ then the maximum value of $\left|z^i\right|$ is

• A.
• B. 1
• C.
• D.

Answer 1

The correct option is D

Let $z=r{e}^{i\theta },-\pi \le \theta <\pi$
$z^i={\left(r{e}^{i\theta }\right)}^i=r^i$. $e^{-\theta }$
= $\left(e^{ln{r}^i}\right)e^{-\theta }=e^{ilnr}.e^{-\theta }$
= (cos ln r + i sin ln r)$e^{-\theta }$
$\therefore \left|z^i\right|=e^{-\theta }\le e^{\pi }$
Since $\theta \ge -\pi ,-\theta \le$ $\pi$, $e^{-\theta }\le e^{\pi }$