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Question: If \(z_{1}\) and \({\bar{z}}_{1}\) represent adjacent vertices of a regular polygon of \(n\) sides a...

If z1z_{1} and zˉ1{\bar{z}}_{1} represent adjacent vertices of a regular polygon of nn sides and if Im(z1)Re(z1)=2−1\frac{{Im}\left( z_{1} \right)}{{Re}\left( z_{1} \right)} = \sqrt{2} - 1, then nn is equal to

A

8

B

16

C

18

D

24

Answer

8

Explanation

Solution

Sol. Since z1z_{1} and zˉ1{\bar{z}}_{1} are the adjacent vertices of a regular polygon of n sides, we have ∠z1ozˉ1=2πn\angle z_{1}o{\bar{z}}_{1} = \frac{2\pi}{n} and

∣z1∣=∣zˉ1∣.|z_{1}| = |{\bar{z}}_{1}|. Thus, z1=zˉ1e2πi/nz_{1} = {\bar{z}}_{1}e^{2\pi i/n}

Let z1=r(cos⁡θ+isin⁡θ)=reiθz_{1} = r\left( \cos\theta + i\sin\theta \right) = re^{i\theta}

⇒ zˉ1=re−iθ{\bar{z}}_{1} = re^{- i\theta}

since z1=zˉ1e2πi/nz_{1} = {\bar{z}}_{1}e^{2\pi i/n}

⇒ reiθ=re−iθe2πi/n=re2πi/n−iθre^{i\theta} = re^{- i\theta}e^{2\pi i/n} = re^{2\pi i/n - i\theta}

⇒ θ=2πn−θ−θ\theta = \frac{2\pi}{n} - \theta - \theta or θ=πn\theta = \frac{\pi}{n}

Therefore z1=r(cos⁡θ+isin⁡θ)z_{1} = r\left( \cos\theta + i\sin\theta \right)=

r[cos⁡(πn)+isin⁡(πn)]r\left\lbrack \cos\left( \frac{\pi}{n} \right) + i\sin\left( \frac{\pi}{n} \right) \right\rbrack

Now, Im(z1)Re(z1)=2−1\frac{{Im}\left( z_{1} \right)}{{Re}\left( z_{1} \right)} = \sqrt{2} - 1 ⇒ rsin⁡(πn)rcos⁡(πn)\frac{r\sin\left( \frac{\pi}{n} \right)}{r\cos\left( \frac{\pi}{n} \right)}=2−1\sqrt{2} - 1

⇒ tan⁡(πn)=2−1\tan\left( \frac{\pi}{n} \right) = \sqrt{2} - 1 = tan(π8)\left( \frac{\pi}{8} \right)

⇒ n = 8