<s>[INST] Let z be a complex number such that $\frac{z-i}{z-1}$ is purely imaginary. Then the minimum value of $|z-(2+2 i)|$ Let
z be a complex number such that $\frac{z-i}{z-1}$ is purely imaginary. Then the minimum value of $|z-(2+2 i)|$ is ?
My approach:-
$$
\begin{array}{l}
 \\
\left|z-z^{\prime}\right| \geqslant| \ |z|-\left|z^{\prime}\right||
\end{array}
$$
where, $z^{\prime}=2+2 i$
$$
\left|z^{\prime}\right|=\sqrt{8}=2 \sqrt{2}
$$
$Let, \quad z=x+i y$
$$
\begin{array}{l}
\frac{z-i}{z-1} \\
=\frac{\{x+i(y-1)\}\{(x-1)-i y\}}{(x-1)^{2}+(i y)^{2}} \\
=\frac{x(x-1)+y(y-1)+i(x-1)(y-1)-x y}{(x-1)^{2}-(i y)^{2}}
\end{array}
$$
${}{} \frac{z-i}{z-1}$ is purely imagivery
$$
\begin{array}{c}
\text { Hence, } x(x-1)+y(y-1)=0 \\
x(x-1)=-y(y-1) \\
x^{2}+y^{2}=x+y
\end{array}
$$
What to do next?
 [/INST] A geometric approach:
$\frac{z-i}{z-1}$ is purely imaginary signifies that $$\arg\left(\frac{z-i}{z-1}\right)={\pi \over 2}+k\pi$$ with $k\in \mathbb{Z}.$
Geometrically, locus of point $z$ is the circle with diameter joining $i$ and $1,$ except these two points.
The center is ${1\over2}+i{1\over2},$ the midpoint of the diameter.
The nearest point of this circle to $2+2i$ lies on the line joining $2+2i$ and the center of the circle.
Therefore, the nearest point is $1+i,$ and the minimal distance $\sqrt2.$

</s>