Question

Let $z=\frac{(2 \sqrt{3}+2 i)^{8}}{(1-i)^{6}}+\frac{(1+i)^{6}}{(2 \sqrt{3}-2 i)^{8}}$. Then argument of $z$ is

• A. $\frac{5 \pi}{6}$
• B. $\frac{\pi}{6}$
• C. $-\frac{\pi}{6}$
• D. $-\frac{5 \pi}{6}$

Answer 1

Answer: (A)

$z=\frac{z_{1}^{8}}{\bar{z}_{2}^{6}}+\frac{z_{2}^{6}}{\bar{z}_{1}^{8}}=\frac{\left|z_{1}\right|^{16}+\left|z_{2}\right|^{12}}{\left(\bar{z}_{2}\right)^{6}\left(\bar{z}_{1}\right)^{8}}$
$\therefore \arg (z)=-8 \arg \left(\bar{z}_{1}\right)-6 \arg \left(\bar{z}_{2}\right)+2 k \pi, k \in I$
$=8 \arg \left(z_{1}\right)+6 \arg \left(z_{2}\right)+2 k \pi$
$=8 \cdot \frac{\pi}{6}+6 \cdot \frac{\pi}{4}+2 k \pi$
$=\frac{4 \pi}{3}+\frac{3 \pi}{2}-2 \pi=\frac{5 \pi}{6}$