Question

If $z(2-2\sqrt{3i}{)}^2=i{(\sqrt{3}+i)}^4$, then arg$\left(z\right)=$

• A. $\frac{5\pi }{6}$
• B. $-\frac{\pi }{6}$
• C. $\frac{\pi }{6}$
• D. $\frac{7\pi }{6}$

Answer 1

Answer: (B)

$-\frac{\pi }{6}$

$i=cis\left(\frac{\pi }{2}\right)$

$(\sqrt{3}+i{)}^4=2^4{(\frac{\sqrt{3}}{2}+\frac{i}{2})}^4$

$=16cis\left(\frac{2i\pi }{3}\right)$

$(2-2\sqrt{3}i{)}^2=4cis\left(\frac{-\pi }{3}\right)$

Hence,

$z=\frac{16cis(\frac{2\pi }{3}+\frac{\pi }{2})}{4cis\left(\frac{-2\pi }{3}\right)}$

$z=4cis\left(\frac{11\pi }{6}\right)$

$=4cis\left(\frac{-\pi }{6}\right)$