Irish Mathematical Olympiad

Let $z$ stand for an arbitrary complex number and denote by $\operatorname{Im}(z)$ its imaginary part. Then we want to minimize the expression
$$f(z) = |\operatorname{Im}(z)|^2 + \sum_{k=1}^{n} |z - a_k|^2.$$
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,
$$f(z) = |\operatorname{Im}(z)|^2 + n|z - m|^2 + \sum_{k=1}^{n} |m - a_k|^2.$$
But, if $m = a + ib$, $z = x + iy$, where $a, b, x, y$ are real numbers, then
$$|\operatorname{Im}(z)|^2 + n|z - m|^2 = y^2 + n[(x - a)^2 + (y - b)^2] = n(x - a)^2 + y^2 + n(y - b)^2,$$
which takes its minimum when $x = a, y = nb/(n + 1)$. Hence
$$$$
$\min$ f = $\text{n 2}$ b^2}{(n+1)^2} + $\text{n}$ 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.