જો $z = \sin \theta - i \cos \theta$ હોય, તો કોઈપણ પૂર્ણાંક $n$ માટે

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