Q49 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

If $z=x+i y, x^2+y^2=1$ and $z_1=z e^{i \theta}$ then $\frac{z_1^{2 n}-1}{z_1^{2 n}+1}=$

  • A
    $-i \tan n\left(\theta+\tan ^{-1}\left(\frac{y}{x}\right)\right)$
  • B
    $i \cot \left(n\left(\theta+\tan ^{-1} \frac{y}{x}\right)\right)$
  • C
    $i \tan n\left(\theta+\tan ^{-1} \frac{x}{y}\right)$
  • Correct answer
    D
    $i \tan \left(n\left(\theta+\tan ^{-1} \frac{y}{x}\right)\right)$

Solution

$z=x+i y=e^{i \phi}$ $\begin{aligned} & z_1=z e^{i \theta}=e^{i(\phi+\theta)} \\ & \frac{z_1^{2 n}-1}{z_1^{2 n}+1}=\frac{\cos (2 n(\theta+\phi)-1+i \sin (2 n(\theta+\phi))}{1+\cos (2 n(\theta+\phi)+i \sin (2 n(\theta+\phi))} \\ & =\frac{2 i \sin n(\theta+\phi) \cos \left(n(\theta+\phi)-2 \sin ^2(n(\theta+\phi))\right.}{2 \cos ^2(n(\theta+\phi)+2 i \sin n(\theta+\phi) \cos n(\theta+\phi)} \\ & =i \tan n(\theta+\phi)=i \tan n\left(\theta+\tan ^{-1} \frac{y}{x}\right)\end{aligned}$

Hence, the correct answer is option: 'D'.

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