Question

Given $\left|z\right|=4$ and $Argz=\frac{5z}{6}$, then $z$ is

• A. $2\sqrt{3}+2i$
• B. $2\sqrt{3}-2i$
• C. $-2\sqrt{3}+2i$
• D. $-\sqrt{3}+i$

Answer 1

The correct option is C $-2\sqrt{3}+2i$

Given that

$\Rightarrow |z|=4$ and $Argz=\frac{5\pi }{6}$

$\Rightarrow$Let $z=x+iy$

$\Rightarrow |z|=4$

$\Rightarrow \sqrt{x^2+y^2}=4$

$\Rightarrow x^2+y^2=16$ ...$\left(1\right)$

$\Rightarrow Argz=\frac{5\pi }{6}$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)=\frac{5\pi }{6}$

$\Rightarrow \frac{y}{x}=\tan \left(\frac{5\pi }{6}\right)$

$\Rightarrow \frac{y}{x}=-\frac{1}{\sqrt{3}}$

$\Rightarrow x^2=3y^2$

Substituting this in $\left(1\right)$ we get,

$\Rightarrow x^2+\frac{x^2}{3}=16$

$\Rightarrow \frac{4x^2}{3}=16$

$\Rightarrow x^2=12$

$\Rightarrow x=\pm 2\sqrt{3}$

$\Rightarrow y=\mp 2$

$\Rightarrow \theta$ lies in $2^{nd}Quadrant$

Hence, $\Rightarrow z=-2\sqrt{3}+2i$