Question

If $z=\cos \theta +i\sin \theta$ then the value of $\frac{z^{2n-1}}{z^{2n+1}}$

• A. $cos2\theta -i\sin 2\theta$
• B. $cosn\theta +isinn\theta$
• C. $i\cot \theta$
• D. $-i\tan \theta$

Answer 1

The correct option is A $cos2\theta -i\sin 2\theta$

$z=cos\theta +isin\theta =e^{i\theta }$.

Now,

$\frac{z^{2n-1}}{z^{2n+1}}=z^{2n-1-2n-1}=z^{-2}=(e^{i\theta }{)}^{-2}=e^{-2i\theta }=cos2\theta -isin2\theta$