Question

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

$z=\frac{z_1^8}{{\bar{z}}_2^6}+\frac{z_2^6}{{\bar{z}}_1^8}=\frac{{|z_1|}^{16}+{|z_2|}^{12}}{{\left({\bar{z}}_2\right)}^6{\left({\bar{z}}_1\right)}^8}$
$\therefore \arg (z)=-8\arg \left({\bar{z}}_1\right)-6\arg \left({\bar{z}}_2\right)+2k\pi ,k\in I$
$=8\arg \left(z_1\right)+6\arg \left(z_2\right)+2k\pi$
$=8\cdot \frac{\pi }{6}+6\cdot \frac{\pi }{4}+2k\pi$
$=\frac{4\pi }{3}+\frac{3\pi }{2}-2\pi =\frac{5\pi }{6}$