Mathematics - Complex Number Question with Solution | TestHub

$\because \frac{z-i}{z-1}$ is purely imaginary.

$\arg \left(\frac{z-i}{z-1}\right)=\frac{\pi }{2}+n\pi ;n\in I$

Geometrically, the locus of $z$ is the circle with end points of diameter is $\left(1,0\right) \& \left(0,1\right)$ and centre is $\left(\frac{1}{2},\frac{1}{2}\right)$.So the nearest point to $\left(3,3\right)$ on the circle is $A\left(1,1\right)$.

Minimum distance is $AP=OP-OA$

$=\sqrt{3^2+3^2}-\sqrt{1^2+1^2}$

$=3\sqrt{2}-\sqrt{2}$

$=2\sqrt{2}$