find the minimum value given constraints

If $${z-i\over z-1}=ir$$ then $$z={i-ir\over 1-ir}$$ where $r$ is real number. Now we get $$|z-2-2i| =\sqrt{5r^2+6r+5\over r^2+1}$$

So you have to calculate the minimum of $$f(r) ={5r^2+6r+5\over r^2+1} = 5+{6r\over r^2+1}$$

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Any way, for a fixed $r$ from a starting formula ${z-i\over z-1}=ir$ we get $${|z-i|\over |z-1|}=|r|$$ so $z$ is on Apollonius circle with respect of points $1$ and $i$ with ratio $|r|$.

If $\frac{z-i}{z-1}$ is a purely imaginary value, then $z$ must lie (in the complex plane) at a point such that the lines from $z$ to $i$ and from $z$ to $1$ are at right angles to one another. This locus is a circle* (shown in blue below):

In the diagram above, $2+2i$ is the point in green. It should be evident that the closest point on the blue circle from $2+2i$ is at $1+i$. The orange circle represents all points that are at distance $\sqrt{2}$ from $2+2i$; the blue circle lies entirely outside the orange circle, save at $1+i$, where they are tangent.

That distance, $\sqrt{2}$, is the minimum value of $|z - (2+2i)|$.

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*This property is known in elementary geometry as Thales's Theorem (among other things). The Wikipedia plot summary for this theorem (linked above) gives a couple of proofs.

$$ \mbox{Re}\left(\frac{z-i}{z-1}\right) = 0\Rightarrow \frac{x-y-1}{(x-1)^2+y^2}+1 = 0 $$

which represents a circle here called $C_1$

now $f(x,y)=\vert z-(2+2i) \vert^2 = 8 - 4 x + x^2 - 4 y + y^2$ which is another circle here called $C_2$

The intersections $C_1$ and $C_2$ gives $(0,0)$ and $(1,1)$

so the minimum value is $\sqrt{2}$ for $(1,1)$