Question

If $z_1$ and ${\bar{z}}_1$ represent adjacent vertices of a regular polygon of $n$ sides and if $\frac{\text{Im}\left(z_1\right)}{\text{Re}\left(z_1\right)}=\sqrt{2}-1$, then $n$ is equal to

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

Answer 1

Answer: (B)

Since $z_1$ and ${\bar{z}}_1$ are the adjacent vertices of a regular polygon of $n$ sides,
we have, $\angle z_{1}0{\bar{z}}_1=\frac{2\pi }{n}$
and, $\left|z_1\right|=\left|{\bar{z}}_1\right|$
Thus, $z_1={\bar{z}}_1{e}^2{\pi }^{\prime /n}$
Let $z_1=r(\cos \theta +i\sin \theta )=r{e}^i\theta$
$\Rightarrow {\bar{z}}_1=r{e}^{-i}\theta$
Since $z_1={\bar{z}}_1{e}^2{\pi }^{\prime /n}$
$\Rightarrow r{e}^i\theta =r{e}^{-i}\theta e^2{\pi }^{i/n}$
$=r{e}^2{\pi }^{i/n-i}\theta$
$\Rightarrow \theta =\frac{2\pi }{n}-\theta$
or $\theta =\frac{\pi }{n}$
Therefore, $z_1=r(\cos \theta +i\sin \theta )$
$=r\left[\cos \frac{\pi }{n}+i\sin \frac{\pi }{n}\right]$
Now, $\frac{\text{Im}\left(z_1\right)}{\text{Re}\left(z_1\right)}=\sqrt{2}-1$
$\Rightarrow \frac{r\sin \left(\frac{\pi }{n}\right)}{r\cos \left(\frac{\pi }{n}\right)}=\sqrt{2}-1$
$\Rightarrow \tan \frac{\pi }{n}=\sqrt{2}-1=\tan \frac{\pi }{8}$
$\Rightarrow n=8$