Question

If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary, then the minimum value of $|z–(3+3i)|$ is

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

Answer 1

The correct option is B $2\sqrt{2}$

$\because \frac{z-i}{z-1}$ is purely imaginary number
$\therefore \frac{z-i}{z-1}+\frac{\overline{z}+i}{\overline{z}-1}=0$
$\Rightarrow (z-i)(\overline{z}-1)+(z-1)(\overline{z}+i)=0$
$\Rightarrow 2z\overline{z}-(z+\overline{z})+i(z-\overline{z})=0$
Let $z=x+iy$
$2(x+iy)(x-iy)-(x+iy+x-iy)+i(x+iy-x+iy)=0$
$\therefore x^2+y^2-x-y=0$
$\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2=\frac{1}{2}$
Which is a circle of centre $(\frac{1}{2},\frac{1}{2})$ and radius $\frac{1}{\sqrt{2}}$
$\therefore$ Minimum value of $|z-(3+3i)|=QR$



$=\sqrt{{(3-\frac{1}{2})}^2+{(3-\frac{1}{2})}^2}-\frac{1}{\sqrt{2}}$
$=\frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2\sqrt{2}$