Question

A point $z$ has been potted in the complex plane, as shown in figure below



$\frac{1}{Z}$ lies in the curve

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let $Z=x+jy$
As $Z$ lies in the first quadrant in the unit circle
$\Rightarrow 0<x<1,0<y<1$ and $0<\sqrt{x^2+y^2}<1$
$\frac{1}{Z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$
Since $x>0,y>0\Rightarrow \frac{x}{x^2+y^2}>0$ and $\frac{-y}{x^2+y^2}<0$
$\Rightarrow \frac{1}{Z}$ lies in IV quardant
$\left|\frac{1}{Z}\right|=\left|\frac{1}{x+iy}\right|=\frac{1}{\sqrt{x^2+y^2}}>1$
$(\because 0<\sqrt{x^2+y^2}<1)$
$\Rightarrow \frac{1}{Z}$ lies outside the unit circle in IV quadrant.