Question

If $\omega =\frac{z}{z-\frac{1}{3}}$ and $|\omega |=1$, then z lies on

• A. a circle
• B. an ellipse
• C. a parabola
• D. a straight line

Answer 1

The correct option is D a straight line

Given $\omega =\frac{3z}{3z-i}\therefore |\omega |=\frac{3|z|}{|3z-i|}$
$\Rightarrow |3z-i|=3|z|$
$\Rightarrow |3\left(x\right)+i(3y-1)|=|3(x+iy)|(z=x+iy)$
$\Rightarrow (3x{)}^2+(3y-1{)}^2=9(x^2+y^2)\Rightarrow 6y-1=0$ which is a straight line.