Question

Let $z=x+iy$ be a complex number where $x,y\in R$ and $i=\sqrt{-1}.$ The area bounded by locus of $P\left(z\right)$ satisfying $|z-1|=2Im\left(z\right)$ and coordinate axes on the complex plane, is equal to
$[$ Here, $Im\left(z\right)$ denotes imaginary part of $z$ $]$

• A. $\sqrt{3}$ sq. unit
• B. $2\sqrt{3}$ sq. unit
• C. $\frac{1}{\sqrt{3}}$ sq. unit
• D. $\frac{1}{2\sqrt{3}}$ sq. unit

Answer 1

The correct option is D $\frac{1}{2\sqrt{3}}$ sq. unit

Given $|z-1|=2Im\left(z\right)$
$\Rightarrow (x-1{)}^2+y^2=4y^2$
$\Rightarrow (x-1{)}^2=3y^2$
$\Rightarrow (x-1)=\pm \sqrt{3}y$
$\therefore L_1:x-\sqrt{3}y-1=0\left[\text{Rejected as}Im\right(z)\ge 0]$
and $L_2:x+\sqrt{3}y-1=0$
Also, $L_3:x=0$ and $L_4:y=0$


So, area of triangle formed by $L_2,L_3$ and $L_4$ is $(\frac{1}{2}\times 1\times \frac{1}{\sqrt{3}})=\frac{1}{2\sqrt{3}}$ sq. unit