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Q.
If $w=\frac{z}{z-\frac{1}{2}i}$ and $\left|w\right|=1,$ then $z$ lies on :
AIEEEAIEEE 2005Complex Numbers and Quadratic Equations
Solution:
Key Idea : If $\left|\frac{z-z_{1}}{z-z_{2}}\right|=k$, then
$z$ lies on a circle of $k \ne1$ 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 $\bot$ 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.