<s>[INST] find the minimum value given constraints 
Let $z$ be a complex number such that $\dfrac{z-i}{z-1}$ is purely imaginary. Find the minimum value of $|z-(2+2i)|$.
Source: ISI 2017 BMATH UGA
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Attempt:
Since $\dfrac{z-i}{z-1}$ is purely imaginary,
$$$$$$\dfrac{z-i}{z-1}+\overline{\left(\dfrac{z-i}{z-1}\right)}=0$$
This reduces to
$$|z|^2=\Re(z)+\Im(z)$$
This represents the locus of $z$ on the Argand Plane. The minimum value of $|z-(2+2i)|$ will be the shortest distance between any point $z$ lying on $|z|^2=\Re(z)+\Im(z)$ and the point $(2,2)$ on the Argand Plane.
$$$$Unable to recognize the locus represented by $|z|^2=\Re(z)+\Im(z)$.
How to identify locus represented by $|z|^2=\Re(z)+\Im(z)$  without reducing to Cartesian Coordinates?
 [/INST] We can just use the triangle inequality and following facts:


*

*$|z-(2+2i)| \geq ||z| - 2\sqrt{2}|$

*$\frac{z-i}{z-1}= ia \Rightarrow z = \frac{i-ia}{1-ia} \Rightarrow |z| = \frac{|1-a|}{\sqrt{1+a^2}}$

*$|z|^2 = \frac{(1-a)^2}{1+a^2} \leq 2$, because
$$\frac{(1-a)^2}{1+a^2} \leq 2 \Leftrightarrow a^2-2a+1 \leq 2a^2 + 2 \Leftrightarrow  0 \leq a^2 + 2a + 1 = (a+1)^2$$


All together yields
$$|z-(2+2i)| \geq ||z| - 2\sqrt{2}| \geq |\sqrt{2} - 2\sqrt{2}| = \sqrt{2} \mbox{ where equality is reached for } a = -1$$
Indeed for $z = \frac{i-i(-1)}{1-i(-1)} = \frac{2i}{1+i} = 1+i$
$$|z - 2(1+i)| = |1+i - 2(1+i)| = |1+i| = \sqrt{2}$$
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