Question

If $w=\frac{z}{z-\frac{1}{2}i}$ and $\left|w\right|=1,$ then $z$ lies on :

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

Answer 1

Answer: (B)

Key Idea : If $\left|\frac{z-z_1}{z-z_2}\right|=k$, then
$z$ lies on a circle of $k\ne 1$ and $z$ lies on a perpendicular bisector of segment with end points $Z_1$ and $z_2$, if $k=1$.
Given that $w=\frac{z}{z-\frac{i}{3}}$ and $\left|w\right|=1$
$\Rightarrow \left|\frac{z}{z-\frac{i}{3}}\right|=1\Rightarrow \left|z\right|=\left|z-\frac{i}{3}\right|$
$\Rightarrow z$ lies on $\perp$ bisector of $\left(0,0\right)$ and $\left(0,\frac{1}{3}\right).$
So, $z$ lies on a straight line.
Alternate Solution
$\because w=\frac{z}{z-\frac{i}{3}}$ and $\left|w\right|=1$
$\Rightarrow \left|\frac{z}{z-\frac{i}{3}}\right|=1$
$\Rightarrow 3\left|z\right|=\left|3z-i\right|$
Let $z=x+iy$
$\therefore 3\left|x+iy\right|=\left|3\left(x+iy\right)-i\right|$
$\Rightarrow 3\sqrt{\left(x^2+y^2\right)}=\sqrt{{\left(3x\right)}^2+{\left(3y-1\right)}^2}$
$\Rightarrow 9x^2+9y^2=9x^2+9y^2+1-6y$
$\Rightarrow y=\frac{1}{6}$
$\therefore$ Which shows that $z$ lies on a straight line.