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} $
$\Rightarrow n=8$