Question

If $P$ is the affix of $z$ in the Argand diagram and $P$ moves so that $\frac{z-i}{z-1}$ is always purely imaginary, then the locus of $z$ is a circle of centre $(\frac{1}{2},\frac{1}{2})$ radius $\frac{1}{\sqrt{2}}.$

• A. True
• B. False

Answer 1

The correct option is B False

$\frac{z-i}{z-1}$

$\frac{z-i}{z-1}\times \frac{z+1}{z+1}$

$\frac{(z-i)(z+1)}{(z-1)(z+1)}$

$\frac{z^2+z-iz-i}{z^2-1^2}$

$\frac{(z^2+z)-i(z+1)}{z^2-1}$

Since $\frac{z-i}{z-1}$ is purely imaginary, means $Re\left(z\right)=0$

$\therefore \frac{z^2+z}{z^2-1}=0$

$\frac{z(z+1)}{(z+1)(z-1)}=0$

$\frac{z}{z-1}=0$

$z\ne 0\Rightarrow z-1=0$

$\therefore z=1$

$|z|=\sqrt{x^2+y^2}=1$

Squaring on both sides, we get,

$x^2+y^2=1\Rightarrow C(0,0),r=1$