Minimum of $\sum\limits_{k=1}^n|z^2_k+1|$ given $\sum\limits_{k=1}^nz_k=0$ Let $n$ be a positive integer and $z_1,z_2,\cdots,z_n$ be  complex numbers such that $$z_1+z_2+\cdots+z_n=0.$$

Problem. Find the minimum of$$
|z^2_1+1|+|z^2_2+1|+\cdots+|z^2_n+1|.$$

I considered that
$$\sum_{i=1}^{n}|z^2_{i}+1|\ge\left|\sum_{i=1}^{n}z^2_{i}+1\right|,$$
then?
 A: As Paolo Leonetti has pointed out, for even $n$, the minimum $0$ is attained by splitting the $z_k$s into pairs of $i,-i$. So, we suppose $n$ is odd in the sequel.
Since the $z_k$s sum to zero, if one of them has a positive real part, some other one must have a negative real part. Suppose $\Re(z_1)>0>\Re(z_2)$. When $t>0$ is sufficiently small, we have
\begin{cases}
|z_1+i|>|z_1-t+i|,\\
|z_1-i|>|z_1-t-i|,\\
|z_2+i|>|z_2+t+i|,\\
|z_2-i|>|z_2+t+i|,
\end{cases}
because the imaginary parts remain intact on the RHS but the sizes of the real parts diminish. However, $|z^2+1|\equiv|z+i||z-i|$. So, if we "pinch" the real parts of $z_1$ and $z_2$ continually towards the origin at the same speed, $|z_1^2+1|+|z_2^2+1|$ will become smaller and smaller until one of the real parts becomes zero. Apply the same procedure to other pairs of $z_k$s with opposite signs of real parts, we conclude that the objective function is minimised only if every $z_k$ is purely imaginary. Put $z_k=iy_k$, the minimisation problem reduces to
$$
\text{minimise } \sum_k|1-y_k^2| \ \text{ subject to } \sum_ky_k=0 \text{ and } \mathbf y=(y_1,\ldots,y_n)\in\mathbb R^n.
$$
Let $A=(0,1)$ and $B=(1,\infty)$. So, $\mathbb R=-B\cup\{-1\}\cup-A\cup\{0\}\cup A\cup\{1\}\cup B$. Using a similar argument to the above, we can always strictly lower the objective function value in each case below:


*

*$y_i\in-B,\ y_j\in-A\cup\{0\}$: pinch-in towards $-1$.

*$y_i\in-B,\ y_j\in A\cup\{1\}$: pinch-in towards the boundaries of $[-1,0]$. Note that $|(y_i+t)^2-1| + |1-(y_j-t)^2|=2t(y_i+y_j)$ plus a constant when $t>0$ is small.

*$y_i\in-B,\ y_j\in B$: pinch-in towards the boundaries of $[-1,1]$.

*$y_i\in-A,\ y_j\in A\cup\{0\}$: spread out towards the boundaries of $[-1,1]$.

*$y_i,y_j\in-A$: spread out towards the boundaries of $[-1,0]$. Note that when $y_i\le y_j$ and $f(t)=\left[1-(y_i-t)^2+1-(y_j+t)^2\right]$, we have $f'(t)=-2(y_j-y_i+2t)$. Hence $f'(0)=-2(y_j-y_i)\le0$ and $f''(0)=-4<0$.


It follows from cases 1 to 3 that, if some element $y_i$ of a global minimiser belongs to $-B$, no $y_j$ may belong to $-A\cup\{0\}\cup A\cup\{1\}\cup B$. In other words, all $y_j$s belong to $-B\cup\{-1\}$. But this is impossible because $\sum_ky_k=0$. Therefore, no element of $\mathbf y$ belongs to $-B$. By symmetry, no element of $\mathbf y$ belongs to $B$ too.
Hence every $\mathbf y\in[-1,1]$.
Now, if there is some $y_i\in-A$, then by the result of case 4, no $y_j$ belongs to $A\cup\{0\}$. Therefore every $y_j\in\{-1\}\cup-A\cup\{1\}$. However, by the result of case 5, $y_i$ must be the only element in $-A$. Hence all other elements are $\pm1$s. Yet, this is impossible because the $y_k$s sum to zero. So, there must be no $y_i$ in $-A$. By symmetry, there is no $y_i$ in $A$ too.
Hence every $y_j\in\{-1,0,1\}$. Since the $y_j$s sum to zero, $-1$ and $1$ must occur in pairs. It is now clear that the minimum occurs when exactly one $y_j$ is zero. Translating back, exactly one $z_j$ is zero and the rest are pairs of $-i,i$. The minimum value is $1$.
