If ${z}_{1}$ and ${z}_{2}$ are two non-zero complex numbers such that $\left| \cfrac { { z }_{ 1 } }{ { z }_{ 2 } } \right| =2$ and $arg(\left( { z }_{ 1 }{ z }_{ 2 } \right) =\cfrac { 3\pi }{ 2 }$, then $\cfrac { \bar { { z }_{ 1 } } }{ { z }_{ 2 } }$ is equal
• A
$2$
• A
$2$
• B
$-2$
• B
$-2$
• C
$-2i$
• C
$-2i$
• D
$2i$
• D
$2i$

#### Solution

Let $z_{1}=a+ib$ and $z_{2}=c+id$
$\left | \dfrac{z_{1}}{z_{2}} \right |=2$
$\left |z _{1} \right |=2\left |z _{2} \right |$
arg($z_{1}z_{2})=\dfrac{3\pi }{2}$
$\tan^{-1}\dfrac{b}{a}+\tan^{-1}\dfrac{d}{c}=\dfrac{3\pi }{2}$
$\tan^{-1}\dfrac{bd+ac}{ad-bc}=\dfrac{3\pi }{2}$
so,$ad=bc$
$\dfrac{b}{a}=\dfrac{d}{c}$
from $\left |z _{1} \right |=2\left |z _{2} \right |$
$a^{2}+b^{2}=4(c^{2}+d^{2})$
$\dfrac{a}{c}$=2
so $\dfrac{z_{1}}{z_{2}}=\dfrac{a+ib}{c+id}=(\dfrac{a}{c})\dfrac{1+i\dfrac{b}{a}}{1+i\dfrac{c}{d}}$
$\dfrac{a}{c}$=2
so option$'A'$ is correct.