Ireland

Let $z$ stand for an arbitrary complex number and denote by $\mathrm{Im}(z)$ its imaginary part. Then we want to minimize the expression $$f(z)=|\mathrm{Im}(z){|}^2+\sum _{k=1}^n|z-a_k{|}^2.$$ If $m$ denotes the centroid of the given points, so that $m=\frac{1}{n}{\sum }_{k=1}^n{a}_k$, then, using $|w{|}^2=w\bar{w}$, we easily obtain $$\sum _{k=1}^n|z-a_k{|}^2=n|z-m{|}^2+\sum _{k=1}^n|m-a_k{|}^2.$$ Hence, $$f(z)=|\mathrm{Im}(z){|}^2+n|z-m{|}^2+\sum _{k=1}^n|m-a_k{|}^2.$$ But, if $m=a+ib$, $z=x+iy$, where $a,b,x,y$ are real numbers, then $$|\mathrm{Im}(z){|}^2+n|z-m{|}^2=y^2+n[(x-a{)}^2+(y-b{)}^2]=n(x-a{)}^2+y^2+n(y-b{)}^2,$$ which takes its minimum when $x=a,y=nb/(n+1)$. Hence $$ \min f = \frac{n^2 b^2}{(n+1)^2} + \frac{n b^2}{(n+1)^2} + \sum_{k=1}^{n} |m - a_k|^2 = \frac{n(\operatorname{Im}(m))^2}{n+1} + \sum_{k=1}^{n} |m - a_k|^2.