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)
$=\frac{2 \cos ^{2} 2 \alpha+2 i \sin 2 \alpha \cos 2 \alpha}{2 \sin ^{2} \frac{3 \alpha}{2}-2 i \sin \frac{3 \alpha}{2} \cos \frac{3 \alpha}{2}}$
$=\frac{2 \cos 2 \alpha}{2 \sin \frac{3 \alpha}{2}} \times \frac{(\cos 2 \alpha+i \sin 2 \alpha)}{\left(\sin \frac{3 \alpha}{2}-i \cos \frac{3 \alpha}{2}\right)}$
$\therefore \left|\frac{1+z^{4}}{1-z^{3}}\right|=\frac{\cos 2 \alpha}{\sin \frac{3 \alpha}{2}}\{$ as $|\cos \,\theta+i \sin \,\theta|=1\}$