Q109 / 155 · WBJEE 2016 · Mathematics

If $z=\sin \theta-i \cos \theta,$ then for any integer $n$

  • Correct answer
    A
    $z^{n}+\frac{1}{z^{n}}=2 \cos \left(\frac{n \pi}{2}-n \theta\right)$
  • B
    $z^{n}+\frac{1}{z^{n}}=2 \sin \left(\frac{n \pi}{2}-n \theta\right)$
  • Correct answer
    C
    $z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)$
  • D
    $z^{n}-\frac{1}{z^{n}}=2 i \cos \left(\frac{n \pi}{2}-n \theta\right)$

Solution

Given, $z=\sin \theta-i \cos \theta$ $=\cos \left(\theta-\frac{\pi}{2}\right)+i \sin \left(\theta-\frac{\pi}{2}\right)=e^{i(\theta-\frac{\pi}{2}).}$ Now. $z^{n}=e^{i(n\theta-\frac{n\pi}{2}).}=\cos \left(n \theta-\frac{n \pi}{2}\right)+i \sin \left(n \theta-\frac{n \pi}{2}\right)$ $\Rightarrow \frac{1}{z^{n}}=e^{i\left(\frac{n \pi}{2}-n\theta\right)}=\cos \left(n \theta-\frac{n \pi}{2}\right)-i \sin \left(n \theta-\frac{n \pi}{2}\right)$ $\therefore \quad z^{n}+\frac{1}{z^{n}}=2 \cos \left(n \theta-\frac{n \pi}{2}\right)=2 \cos \left(\frac{n \pi}{2}-n \theta\right)$ and $z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)$

Hence, the correct answer is options: 'A', 'C'.

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