Question

If $z=\cos \theta +i\sin \theta$, then

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

Answer 1

Answer: (A), (C)

The correct options are
A $z^n+\frac{1}{z^n}=2\cos n\theta$
C $z^n-\frac{1}{z^n}=2i\sin n\theta$

We have $\frac{1}{z}=\frac{1}{\cos \theta +i\sin \theta }=\cos \theta -i\sin \theta$

$\therefore z^n={(\cos \theta +i\sin \theta )}^n=\cos n\theta +i\sin n\theta$

and $\frac{1}{z^n}={(\cos \theta -i\sin \theta )}^n=\cos n\theta -i\sin n\theta$

Hence $z^n+\frac{1}{z^n}=2\cos n\theta$ and $z^n-\frac{1}{z^n}=2i\sin n\theta$