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 :

• A. $2 \sqrt{2}-1$
• B. $3 \sqrt{2}$
• C. $6 \sqrt{2}$
• D. $2 \sqrt{2}$

Answer 1

Answer: (D)

$\frac{z-i}{z-1}$ is purely Imaginary number
Let $z=x+$ iy
$\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 |z-(3+3 i)|_{\min } =|P C|-\frac{1}{\sqrt{2}} $
$=\frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2 \sqrt{2} $