If z₁ and ${\overline{\mathrm{z}}}_1$ represent adjacent vertices of a regular polygon of n sides and if $\frac{\mathrm{Im}\left({\mathrm{z}}_1\right)}{\mathrm{Re}\left({\mathrm{z}}_1\right)}=\sqrt{2}-1$, then n is equal to

Since z1 and ${\overline{\mathrm{z}}}_1$ are the adjacent vertices of a regular polygon of n sides, we have, $\mathrm{\angle }{\mathrm{z}}_{1}0{\overline{\mathrm{z}}}_1=\frac{2\mathrm{\pi }}{\mathrm{n}}$

and, $\left|{\mathrm{z}}_1\right|=\left|{\overline{\mathrm{z}}}_1\right|$

Thus, ${\mathrm{z}}_1={\overline{\mathrm{z}}}_1{\mathrm{e}}^2{\mathrm{\pi }}^{\mathrm{j}/\mathrm{n}}$

Let ${\mathrm{z}}_1=\mathrm{r}(\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta })={\mathrm{re}}^{\mathrm{i}}\mathrm{\theta }$

$\Rightarrow {\overline{\mathrm{z}}}_1={\mathrm{re}}^{-\mathrm{i}}\mathrm{\theta }$

Since           ${\mathrm{z}}_1={\overline{\mathrm{z}}}_1 {\mathrm{e}}^2{\mathrm{\pi }}^{\mathrm{j}/\mathrm{n}}$

$\begin{array}{l}\Rightarrow {\mathrm{re}}^{\mathrm{i}}\mathrm{\theta }={\mathrm{re}}^{-\mathrm{i}}{\mathrm{\theta e}}^2{\mathrm{\pi }}^{\mathrm{i}/\mathrm{n}}={\mathrm{re}}^2{\mathrm{\pi }}^{\mathrm{i}/\mathrm{n}-\mathrm{i}}\mathrm{\theta }\\ \Rightarrow \mathrm{\theta }=\frac{2\mathrm{\pi }}{\mathrm{n}}-\mathrm{\theta }\text{ or }\mathrm{\theta }=\frac{\mathrm{\pi }}{\mathrm{n}}\end{array}$

Therefore, ${\mathrm{z}}_1=\mathrm{r}(\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta })=\mathrm{r}\left[\cos \frac{\mathrm{\pi }}{\mathrm{n}}+\mathrm{isin}\frac{\mathrm{\pi }}{\mathrm{n}}\right]$

Now, $\frac{\mathrm{Im}\left({\mathrm{z}}_1\right)}{\mathrm{Re}\left({\mathrm{z}}_1\right)}=\sqrt{2}-1\Rightarrow \frac{\mathrm{rsin}\left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)}{\mathrm{rcos}\left(\frac{\mathrm{\pi }}{\mathrm{n}}\right)}=\sqrt{2}-1$

$\Rightarrow \tan \frac{\mathrm{\pi }}{\mathrm{n}}=\sqrt{2}-1=\tan \frac{\mathrm{\pi }}{8}\Rightarrow \mathrm{n}=8$