Question

if $z(2-i2\sqrt{3}{)}^2=i(\sqrt{3}+i{)}^4$ then amplitude of $z$ is

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

Answer 1

Answer: (B)

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

Let $w$ be the cube root of unity
$\therefore w^3=1\&1+{w+w}^2=0$
Where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$ ...(1)
$z{(2-2i\sqrt{3})}^2=i{(\sqrt{3}+i)}^4\Rightarrow z{\left(\frac{1-i\sqrt{3}}{2}\right)}^2=i{\left(\frac{\sqrt{3}+i}{2}\right)}^4\Rightarrow z{(-w)}^2=i{\left(\frac{-1+i\sqrt{3}}{2}\right)}^4\Rightarrow z{(-w)}^2=i{w}^4\Rightarrow z=i{w}^2\Rightarrow z=\frac{\sqrt{3}-i}{2}\therefore arg\left(z\right)=-\frac{\pi }{6}$
Hence, option 'B' is correct.