Question

If $z$ satisfies the equation $|z|-z=1+2i$, then $z$ is equal to

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

Answer 1

Answer: (B)

We have, $|z|-z=1+2i$
If $z=x+iy$, then this equation reduces to
$|x+iy|-(x+iy)=1+2i$
$\Rightarrow \left(\sqrt{x^2+y^2}-x\right)+(-iy)=1+2i$
On comparing real and imaginary parts of both sides of this equation, we get $\sqrt{x^2+y^2}-x=1$
$\Rightarrow \sqrt{x^2+y^2}=1+x$
$\Rightarrow x^2+y^2=(1+x{)}^2$
$\Rightarrow x^2+y^2=1+x^2+2x$
$\Rightarrow y^2=1+2x$ ...(i)
and $-y=2$
$\Rightarrow y=-2$
Putting this value in Eq. (i), we get
$(-2{)}^2=1+2x$
$\Rightarrow 2x=3$
$\Rightarrow x=\frac{3}{2}$
Hence, $z=x+iy$
$=\frac{3}{2}-2i$