Question:
If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary, then the minimum value of $|z-(3+3 i)|$ is:
Correct Option: , 4
Solution:
$\frac{z-i}{z-1}$ is purely Imaginary number
Let $z=x+i y$
$\therefore \frac{x+i(y-1)}{(x-1)+i(y)} \times \frac{(x-1)-i y}{(x-1)-i y}$
$\Rightarrow \frac{x(x-1)+y(y-1)+i(-y-x+1)}{(x-1)^{2}+y^{2}}$ is purely
Imaginary number
$\Rightarrow x(x-1)+y(y-1)=0$
$\Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$
$\therefore|\mathrm{z}-(3+3 \mathrm{i})|_{\min }=|\mathrm{PC}|-\frac{1}{\sqrt{2}}$
$=\frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2 \sqrt{2}$