Question

If $z=\cos \alpha +i\sin \alpha ;0<\alpha <\frac{\pi }{4}$, then $\left|\frac{1+z^4}{1-z^3}\right|=$

• A. $\frac{\cos 2\alpha }{\sin \frac{3}{2}\alpha }$
• B. $\frac{\cos \alpha }{\sin \frac{3}{2}\alpha }$
• C. $\frac{\cos 2\alpha }{\sin \frac{\alpha }{2}}$
• D. $\frac{\cos \alpha }{\sin \frac{\alpha }{2}}$

Answer 1

Answer: (A)

It is given that $z=\cos \alpha +i\sin \alpha ,0<\alpha <\frac{\pi }{4}$
So, $\frac{1+z^4}{1-z^3}=\frac{1+(\cos \alpha +i\sin \alpha {)}^4}{1-(\cos \alpha +i\sin \alpha {)}^3}$
$=\frac{1+\cos 4\alpha +i\sin 4\alpha }{1-\cos 3\alpha -i\sin 3\alpha }$ (by De-Moivre's theorem)
$=\frac{2{\cos }^{2}2\alpha +2i\sin 2\alpha \cos 2\alpha }{2{\sin }^2\frac{3\alpha }{2}-2i\sin \frac{3\alpha }{2}\cos \frac{3\alpha }{2}}$
$=\frac{2\cos 2\alpha }{2\sin \frac{3\alpha }{2}}\times \frac{(\cos 2\alpha +i\sin 2\alpha )}{\left(\sin \frac{3\alpha }{2}-i\cos \frac{3\alpha }{2}\right)}$
$\therefore \left|\frac{1+z^4}{1-z^3}\right|=\frac{\cos 2\alpha }{\sin \frac{3\alpha }{2}}\{$ as $|\cos \theta +i\sin \theta |=1\}$