Question

If $z_1$ and $\overline{z_1}$ represent adjacent vertices of a regular polygon of $n$ sides and if $\frac{Im\left(z_1\right)}{Re\left(z_1\right)}=\sqrt{2}-1,$ then $n$ is equal to

• A. $4$
• B. $8$
• C. $16$
• D. None of these

Answer 1

The correct option is B $8$

Since $z_1$ and ${\overline{z}}_1$ are the adjacent vertices of a regular polygon of $n$ sides,

we have, $\angle z_{1}0{\overline{z}}_1=\frac{2\pi }{n}$ and, $\left|z_1\right|=\left|{\overline{z}}_1\right|$

Thus, $z_1={\overline{z}}_1{e}^{\frac{2\pi i}{n}}$

Let $z_1=r(\cos \theta +i\sin \theta )=r{e}^i\theta$

$\Rightarrow {\overline{z}}_1=r{e}^{-i}\theta$

Since $z_1={\overline{z}}_1{e}^2{\pi }^{\frac{i}{n}}$

$\Rightarrow r{e}^i\theta =r{e}^{-i}\theta e^2{\pi }^{\frac{i}{n}}=r{e}^2{\pi }^{\frac{i}{n-i}}\theta$

$\Rightarrow \theta =\frac{2\pi }{n}-\theta$ or $\theta =\frac{\pi }{n}$

Therefore, $z_1=r(\cos \theta +i\sin \theta )=r[\cos \frac{\pi }{n}+i\sin \frac{\pi }{n}]$

Now, $\frac{Im\left(z_1\right)}{Re\left(z_1\right)}=\sqrt{2}-1\Rightarrow \frac{r\sin \left(\frac{\pi }{n}\right)}{r\cos \left(\frac{\pi }{n}\right)}=\sqrt{2}-1$

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

$\Rightarrow n=8$