If z is a complex number such that

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:

  1. $2 \sqrt{2}-1$

  2. $3 \sqrt{2}$

  3. $6 \sqrt{2}$

  4. $2 \sqrt{2}$


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

$\Rightarrow x(x-1)+y(y-1)=0$

$\Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$

$\therefore|\mathrm{z}-(3+3 \mathrm{i})|_{\min }=|\mathrm{PC}|-\frac{1}{\sqrt{2}}$

$=\frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}}=2 \sqrt{2}$

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