Solve the following

Question:

If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}$, then

(a) Re (z) = 0

(b) Im (z) = 0

(c) Re (z) > 0, Im (z) > 0

(d) Re (z) > 0, Im (z) < 0

Solution:

for $\frac{\sqrt{3}}{2}+\frac{i}{2}$

$r=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$ and $\theta=\tan ^{-1} \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\tan ^{-1} \frac{1}{\sqrt{3}}$

i. e. $\theta=\frac{\pi}{6}$

i.e. $\frac{\sqrt{3}}{2}+\frac{i}{2}=r(\cos \theta+i \sin \theta)$

we have $\theta=\frac{\pi}{6}$ and $r=1$

and $\frac{\sqrt{3}}{2}-\frac{i}{2}=r(\cos \theta-i \sin \theta)$

$\therefore z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}$

$=[r(\cos \theta+i \sin \theta)]^{5}+[r(\cos \theta-i \sin \theta)]^{5}$ where $r=1, \theta=\frac{\pi}{6}$

$=(\cos \theta+i \sin \theta)^{5}+(\cos \theta-i \sin \theta)^{5}$

$=\cos 5 \theta+i \sin 5 \theta+\cos 5 \theta-i \sin 5 \theta \quad$ (By De $-$ moivre theorem)

$=2 \cos 5 \theta$