Question

If $z$ is a complex number such that $\frac{z−i}{z−1}$ is purely imaginary, then the minimum value of $âˆ\pounds z−(3+3i)âˆ\pounds$ 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
$∴\frac{x+i(y−1)}{(x−1)+i(y)}×\frac{(x−1)−iy}{(x−1)−iy}$
$⇒\frac{x(x−1)+y(y−1)+i(−y−x+1)}{(x−1{)}^2+y^2}$
is purely Imaginary number
$⇒x(x−1)+y(y−1)=0$
$⇒{\left(x−\frac{1}{2}\right)}^2+{\left(y−\frac{1}{2}\right)}^2=\frac{1}{2}$

$∴âˆ\pounds z−(3+3i){âˆ\pounds }_{\min }=âˆ\pounds PCâˆ\pounds −\frac{1}{\sqrt{2}}$
$=\frac{5}{\sqrt{2}}−\frac{1}{\sqrt{2}}=2\sqrt{2}$