Question

If $z+\frac{1}{z}=2\cos \theta$, where $z$ is complex number and $i=\sqrt{-1}$, then $\frac{z^n-1}{z^n+1}$ is

• A. $\tan \left(\frac{n\theta }{2}\right)$
• B. $i\tan \left(\frac{n\theta }{2}\right)$
• C. $\cot \left(\frac{n\theta }{2}\right)$
• D. $i\cot \left(\frac{n\theta }{2}\right)$

Answer 1

The correct option is B $i\tan \left(\frac{n\theta }{2}\right)$

Let $z=r(\cos \theta +i\sin \theta )$
$\Rightarrow \frac{1}{z}=r(\cos \theta -i\sin \theta )$
$z+\frac{1}{z}=2\cos \theta \Rightarrow r(\cos \theta +i\sin \theta +\cos \theta -i\sin \theta )=2\cos \theta \Rightarrow r=1$
$\therefore z=\cos \theta +i\sin \theta$
Now,
$\frac{z^n-1}{z^n+1}=\frac{\cos n\theta +i\sin n\theta -1}{\cos n\theta +i\sin n\theta +1}=\frac{-2{\sin }^2\frac{n\theta }{2}+2i\sin \frac{n\theta }{2}\cos \frac{n\theta }{2}}{2{\cos }^2\frac{n\theta }{2}+2i\sin \frac{n\theta }{2}\cos \frac{n\theta }{2}}=\tan \left(\frac{n\theta }{2}\right)\left(\frac{-\sin \frac{n\theta }{2}+i\cos \frac{n\theta }{2}}{\cos \frac{n\theta }{2}+i\sin \frac{n\theta }{2}}\right)=\tan \left(\frac{n\theta }{2}\right)\left(\frac{e^{i(\frac{\pi }{2}+\frac{n\theta }{2})}}{e^{\frac{in\theta }{2}}}\right)=\tan \left(\frac{n\theta }{2}\right)e^{i\frac{\pi }{2}}=i\tan \left(\frac{n\theta }{2}\right)$