Question

If $n$ is a positive integer, prove that
$|\mathrm{Im}\left(z^n\right)|\le n|\mathrm{Im}\left(z\right)|{\left|z\right|}^{n-1}$

Answer 1

We have to prove that
$|\mathrm{Im}\left(z^n\right)|\le n|\mathrm{Im}\left(z\right)|{\left|z\right|}^{n-1}$ ... (i)
We know

$\mathrm{Im}\left(z\right)=\frac{z-\overline{z}}{2i}$

$\Rightarrow \mathrm{Im}\left(z^n\right)=\frac{z^n-{\overline{z}}^n}{2i}$

$\therefore Fromequation\left(i\right)$

$\left|\frac{z^n-{\overline{z}}^n}{2i}\right|\le n\left|\frac{z-\overline{z}}{2i}\right||z{|}^{n-1}$

$\Rightarrow \left|\frac{z^n-{\overline{z}}^n}{z-\overline{z}}\right|\le n|z{|}^{n-1}$

Now,

$\left|\frac{z^n-{\overline{z}}^n}{z-\overline{z}}\right|=|z^{n-1}+z^{n-2}\overline{z}+z^{n-3}{\overline{z}}^2+...+{\overline{z}}^{n-1}|$

$\le |z^{n-1}|+|z^{n-2}\overline{z}|+|z^{n-3}{\overline{z}}^2|+...+{\overline{z}}^{n-1}|$

$=|z|n-1+|z|n-2|\overline{z}|+|z{|}^{n-3}|{\overline{z}}^2|+...+|{\overline{z}}^{n-1}|$

$=|z{|}^{n-1}+|z{|}^{n-1}+|z{|}^{n-1}+...+|z{|}^{n-1}$

$(\because |z|=|\overline{z}|)$

$=n|z{|}^{n-1}$

Hence proved.