Question

If $z_1$ and $\overline{z_2}$ 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. $8$
• B. $16$
• C. $18$
• D. $24$

Answer 1

The correct option is A $8$

Since $z_1$ and ${\overline{z}}_2$ 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 $\mid z_1\mid =\mid {\overline{z}}_1\mid .$
Thus, $z_1={\overline{z}}_1{e}^{2\pi i/n}$
Let $z_1=r(cos\theta +isin\theta )=r{e}^{i\theta }$
$\Rightarrow {\overline{z}}_1=r{e}^{-i\theta }$
since $z_1={\overline{z}}_1^{2\pi i/n}$
$\Rightarrow r{e}^{i\theta }=r{e}^{-i\theta }{e}^{2\pi i/n}=r{e}^{2\pi i/n-i\theta }$
$\Rightarrow \theta =\frac{2\pi }{n}-\theta -\theta or\theta =\frac{\pi }{n}$
Therefore $z_1=r(cos\theta +isin\theta )=r\left[cos\right(\frac{\pi }{n})+isin(\frac{\pi }{n}\left)\right]$
Now,$\frac{Im\left(z_1\right)}{Re\left(z_1\right)}=\sqrt{2}-1\Rightarrow \frac{rsin\left(\frac{\pi }{n}\right)}{rcos\left(\frac{\pi }{n}\right)}=\sqrt{2}-1$

$\Rightarrow tan\left(\frac{\pi }{n}\right)=\sqrt{2}-1=tan\frac{\pi }{8}$
$\Rightarrow n=8$