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.