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?
On
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:
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 $1$.
On
Let $n=2m+1$ for some $m\in\mathbb{N}$. Denote by $z:=(z_1,...,z_n)$ and $$S(z):=\sum_{k=1}^n|z_k^2+1|$$ The problem can be expressed as $$\min_{z\in\mathbb{C}}S(z)\hspace{0.2cm}\text{subject to}\hspace{0.2cm}z_1+...+z_n=0$$ Let $\tilde{z}:=(\tilde{z}_1,...,\tilde{z}_n)$ with $\tilde{z}_1=0$ and $\tilde{z}_k=(-1)^ki$ for $k\geqslant 2$. Then $S(\tilde{z})=1$. Clearly if $z^*$ is a minimizer then $S(z^*)\leqslant S(\tilde{z})$.
$\textbf{Observation 1:}$ If $z^*$ is a minimizer then $z^*_k\notin\mathbb{R}\setminus\{0\}$ for every $k$. If there is some $k$ such that $z^*_k\in\mathbb{R}\setminus\{0\}$ then $S(z^*)\geqslant |(z^*_k)^2+1|>1$. Hence a contradiction.
$\textbf{Observation 2:}$ If $z^*$ is a minimizer then at most one $z^*_k=0$. If there are more then one (say two) $z_k^*=0, z^*_l=0$ then $S(z^*)\geqslant |(z^*_k)^2+1|+|(z^*_l)^2+1|=2>1$. Again a contradiction.
$\textbf{Observation 3:}$ Let $z_k:=a_k+ib_k$ where $a_k,b_k\in\mathbb{R}$ for each $k$. If $z$ is a minimizer then $|a_k|\leqslant 1$ and $|b_k|\leqslant2$ for all $k$. That is to say $z_k\in\mathbb{B}(0,\sqrt{5}):=\{z\in\mathbb{C}:|z|\leqslant\sqrt{5}\}$ for each $k$. Now $$|z_k^2+1|^2=|(a_k+ib_k)^2+1|^2=|a_k^2-b_k^2+1+2ia_kb_k|^2\\=(a^2_k-b^2_k+1)^2+4a^2_kb^2_k=(a^2_k+b^2_k+1)^2-4b^2_k\\=(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)$$ If for some $k$ we have $|a_k|>1$ or $|b_k|>2$ then $$S(z)\geqslant |z^2_k+1|=\sqrt{(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)}>1=S(\tilde{z})$$ Therefore such a $z$ would not be a minimum. Clearly for $\tilde{z}$ we have $\tilde{a}_k=0$ for all $k$ and $\tilde{b}_1=0, \tilde{b}_k=(-1)^k$ for all $k$. Surely for the claim to hold it is sufficient to prove that if $z^*$ is a minimizer at least one $z^*_k=0$.
$\textbf{Observation 4:}$ Using the notation $z_k=a_k+ib_k$ the constraint $z_1+...+z_n=0$ is equivalent $\sum_k a_k=0$ and $\sum_kb_k=0$. Let $a:=(a_1,...,a_n), b:=(b_1,...,b_n)$ then the original problem can be rephrased in terms of $a_k,b_k$ as follows $$\min_{z}S(z)=\min_{a,b\in A^n}\sum_{k=1}^n\sqrt{(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)}\hspace{0.2cm}\text{subject to}\hspace{0.2cm}\sum_{k=1}^na_k=\sum_{k=1}^nb_k=0$$ where $A:=\{(x,y)\in\mathbb{R}^2:|x|\leqslant 1, |y|\leqslant 2\}$ (by observation 3 it is sufficient to look at this set $A$). The above problem is just a standard optimization problem of convex kind since constraints, set $A$ and the objective function are convex. Also notice that direct inspection reveals that the minimum cannot occur on the boundary of $A^n$. In either case $|a_k|=1$ or $|b_k|=2$ for some $k$ yields $S(z)>1$. Therefore the minimum must be attained in the interior of $A^n$. The Lagrangian for this problem would be: $$L(a,b,\lambda,\beta):=\sum_{k=1}^n\sqrt{(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)}-\lambda \sum_{k=1}^na_k-\beta \sum_{k=1}^nb_k$$ restricted to $(a,b)\in A^n$ (I am not including this restriction in the Lagrangian because it gets messier). If $(a,b)$ is an optimal solution and the objective function is continuously differentiable there then from the first order conditions we get all $k$: $$\frac{2a_k(a_k^2+b_k^2+1)}{\sqrt{(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)}}=\lambda$$ $$\frac{2b_k(a_k^2+b_k^2-1)}{\sqrt{(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)}}=\beta$$ Notice that by observation $1$ there is no $k$ such that $a_k\neq 0$ and $b_k=0$. Moreover since we are interested in points other than $\tilde{z}$ (or its permutation) then we can also assume that there is no $k$ for which $z_k=0$ that is simultaneously $a_k=b_k=0$. We distinguish now few cases which can be candidates for the first order conditions above:
$\textbf{Case 1:}$ $a_k=a_{k+1}$ and $b_k=b_{k+1}$ for all $k$ which from the constraints implies $a_k=b_k=0$ for all $k$. But this cannot be a minimum since $S(0)=n>1$.
$\textbf{Case 2:}$ If for some $j$ we have $a_j=0$ then necessarily $a_k=0$ for all $k$. If some $b_k=0$ then back to case of $\tilde{z}$. So assume for all $k$ we have $b_k\neq 0$.Then this implies from the second equation for all $k$ we have $$\frac{2b_k(b^2_k-1)}{|b^2_k-1|}=\beta\Rightarrow |b_k|=|b_{k+1}|$$ From the constraint $\sum_kb_k=0$ we must have half of $b_k$ negative and the other half positive but this is impossible since $n=2m+1$ is an odd number.
$\textbf{Case 3:}$ There is some $j$ such $b_j=0$ but then $a_j=0$ and back to $\tilde{z}$.
$\textbf{Case 4:}$ There is some $j$ such that $a^2_j+b^2_j-1=0$ but then for all $k$ we have $a^2_k+b^2_k-1=0$ hence $$\lambda=\frac{2a_k(a_k^2+b_k^2+1)}{\sqrt{(a^2_k+(b_k-1)^2)(a^2_k+(b_k+1)^2)}}=\frac{2a_k}{|a_k|}$$ This implies that all $a_k$ have the same sign and since $a_k\neq 0$ for all $k$ then $\sum_ka_k=0$ cannot hold.
$\textbf{Case 5:}$ When $a^2_k+b^2_k-1>0 \hspace{0.1cm}(<0)$ then $b_k$ must have the same sign for all $k$. Since $b_k\neq 0$ for all $k$ then $\sum_kb_k=0$ cannot hold.
These cases cover all possibilities of interior points being critical points. Only $a_k=b_k=0$ for all $k$ is a critical point but that is not a minimum. On the other hand $S(z)$ is continuous and $A^n$ is a compact set so by Weierstrass theorem $S(z)$ must necessarily attain its minimum on $A^n$. Previous arguments showed that the minimum can be attained in the interior $A^n$ only. Since no interior point where the objective function is continuously differentiable can be a minimum it then forces $\tilde{z}$ (or its permutations) to be a minimum.
$\def\i{\mathrm{i}}$As is pointed out by @PaoloLeonetti, for even $n$, the minimum is $0$. Now suppose $n = 2m + 1$ where $m \in \mathbb{N}$. It will be proved that$$ \sum_{k = 1}^{2m + 1} |z_k^2 + 1| \geqslant 1. $$ The equality can be achieved for $z_1 = 0,\ z_k = (-1)^k \i\ (k \geqslant 2)$.
Lemma: If $a_1, \cdots, a_{2m + 1} \in \mathbb{R}$, $|a_k| \leqslant 1$ for $1 \leqslant k \leqslant 2m + 1$, and $\sum\limits_{k = 1}^{2m + 1} a_k = 0$, then$$ \sum_{k = 1}^{2m + 1} a_k^2 \leqslant 2m. $$
Proof of lemma: If there exists $k_0$ such that $a_{k_0} = 0$, then$$ \sum_{k = 1}^{2m + 1} a_k^2 = \sum_{k \neq k_0} a_k^2 \leqslant \sum_{k \neq k_0} 1^2 = 2m. $$
Now suppose $a_k \neq 0$ for each $k$. Since there are $2m + 1$ nonzero numbers, then either there are at least $m + 1$ positive numbers or there are at least $m + 1$ negative numbers. Without loss of generality, suppose $a_1, \cdots, a_k > 0$, $k \geqslant m + 1$, and $a_1 = \min\limits_{1 \leqslant j \leqslant k} a_k$. Note that$$ \sum_{j = 1}^k a_j = -\sum_{j = k + 1}^{2m + 1} a_j \leqslant 2m + 1 - k \leqslant m. \tag{1} $$
Denote $S = \sum\limits_{j = 1}^{2m + 1} a_j^2$. Take$$a_{1,1} = a_1 - \min(a_1, 1 - a_2),\ a_{1,2} = a_2 + \min(a_1, 1 - a_2),\ a_{1,j} = a_j\ (j \geqslant 3),$$ then\begin{align*} S_1 - S &= \sum_{j = 1}^{2m + 1} a_{1,j}^2 - \sum_{j = 1}^{2m + 1} a_j^2 = (a_{1,1}^2 + a_{1,2}^2) - (a_1^2 + a_2^2)\\ &= 2(a_2 - a_1) \min(a_1, 1 - a_2) + (\min(a_1, 1 - a_2))^2 \geqslant 0, \end{align*} and$$ \sum_{j = 1}^k a_{1,j} = \sum_{j = 1}^k a_j, \quad |a_{1,j}| \leqslant 1\ (1 \leqslant j \leqslant 2m + 1), \quad a_{1,1} = \min_{1 \leqslant j \leqslant k} a_{1, j}.$$
If $a_{1,j} = 0$, then again$$ \sum_{j = 1}^{2m + 1} a_j^2 \leqslant \sum_{j = 1}^{2m + 1} a_{1,j}^2 = \sum_{j > 1} a_{1,j}^2 \leqslant \sum_{j > 1} 1^2 = 2m. $$ Otherwise $a_{1,2} = 1$. Take$$ a_{2,1} = a_{1,1} - \min(a_{1,1}, 1 - a_{1,3}),\ a_{2,3} = a_{1,3} + \min(a_{1,1}, 1 - a_{1,3}),\ a_{2,j} = a_{1,j}\ (j \neq 1, 3), $$ then analogously there is $S_2 \geqslant S_1$ with$$ \sum_{j = 1}^k a_{2,j} = \sum_{j = 1}^k a_{1,j}, \quad |a_{2,j}| \leqslant 1\ (1 \leqslant j \leqslant 2m + 1), \quad a_{2,1} = \min_{1 \leqslant j \leqslant k} a_{2, j}.$$
Now it will be proved that such adjustments can be carried out for at most $k - 1$ times, i.e. there exists $1 \leqslant t \leqslant k - 1$ such that$$ a_1 \geqslant a_{1,1} \geqslant \cdots \geqslant a_{t,1} = 0. $$ If otherwise, then $a_{k - 1, 1} > 0$ and $a_{k - 1, j} = 1$ for $2 \leqslant j \leqslant k$. Thus$$ \sum_{j = 1}^k a_j = \sum_{j = 1}^k a_{k - 1, j} > \sum_{j = 2}^k a_{k - 1, j} \geqslant k - 1 \geqslant m, $$ contradictory to (1). Therefore, suppose $1 \leqslant t \leqslant k - 1$ satisfies$$ a_1 \geqslant a_{1,1} \geqslant \cdots \geqslant a_{t,1} = 0, $$ then$$ \sum_{j = 1}^{2m + 1} a_j^2 \leqslant \sum_{j = 1}^{2m + 1} a_{t,j}^2 = \sum_{j > 1} a_{t,j}^2 \leqslant \sum_{j > 1} 1^2 = 2m. $$
Now back to the question. Suppose $z_k = x_k + \i y_k \ (x_k, y_k \in \mathbb{R})$ for each $k$. If there exists $k$ such that $x_k \neq 0$, without loss of generality suppose $k = 1$ and $x_1 > 0$, then by $\sum z_k = 0$, there exists $l \neq 1$ such that $x_l < 0$. Without loss of generality suppose $l = 2$.
Take$$ x_1' = x_1 - \min(x_1, -x_2) \geqslant 0,\ x_2' = x_2 + \min(x_1, -x_2) \leqslant 0,\ x_k' = x_k \ (k \geqslant 3), $$ and $z_k' = x_k' + \i y_k$ for each $k$, then$$ |z_1 + \i|^2 - |z_1' + \i|^2 = x_1^2 - x_1'^2 > 0 \Longrightarrow |z_1 + \i| > |z_1' + \i|,\\ |z_2 + \i|^2 - |z_2' + \i|^2 = x_2^2 - x_2'^2 > 0 \Longrightarrow |z_2 + \i| > |z_2' + \i|. $$ Analogously,$$ |z_1 - \i| > |z_1' -\i|, \quad |z_2 - \i| > |z_2' -\i|. $$ Therefore,\begin{align*} &\mathrel{\phantom{=}} \sum_{k = 1}^{2m + 1} |z_k^2 + 1| - \sum_{k = 1}^{2m + 1} |z_k'^2 + 1| = (|z_1^2 + 1| + |z_2^2 + 1|) - (|z_1'^2 + 1| + |z_2'^2 + 1|)\\ &= (|z_1 + \i| \cdot |z_1 - \i| - |z_1' + \i| \cdot |z_1' - \i| ) + (|z_2 + \i| \cdot |z_2 - \i| - |z_2' + \i| \cdot |z_2' - \i| ) > 0. \end{align*}
This implies if there exists $k$ such that $x_k \neq 0$, then there is another tuple $(z_1', \cdots, z_{2m + 1}')$ satisfies $\sum z_k' = 0$ and $\sum z_k'^2 < \sum z_k^2$. Therefore,$$ \min_{\substack{z_1, \cdots, z_n \in \mathbb{C}\\\sum z_k = 0}} \sum_{k = 1}^{2m + 1} |z_k^2 + 1| = \min_{\substack{z_1, \cdots, z_n \in \mathbb{C}\\\sum z_k = 0\\\mathrm{Re}(z_k) = 0\ (1 \leqslant k \leqslant 2m + 1)}} \sum_{k = 1}^{2m + 1} |z_k^2 + 1|. $$
Now it suffices to prove that for any $y_1, \cdots, y_{2m + 1} \in \mathbb{R}$, if $\sum\limits_{k = 1}^{2m + 1} y_k = 0$, then$$ \sum_{k = 1}^{2m + 1} |1 - y_k^2| \geqslant 1. $$
If $M = \max\limits_{1 \leqslant k \leqslant 2m + 1} |y_k| > 1$, without loss of generality, suppose $y_1 = M$, then there exists $k$ such that $y_k < 0$. Without loss of generality, suppose $y_2 < 0$. Take$$ y_1' = 1,\ y_2' = y_2 + y_1 - 1, y_k' = y_k\ (k \geqslant 3), $$ then\begin{align*} &\mathrel{\phantom{=}} \sum_{k = 1}^{2m + 1} |1 - y_k^2| - \sum_{k = 1}^{2m + 1} |1 - y_k'^2| = |1 - y_1^2| + |1 - y_2^2| - |1 - y_2'^2|\\ &= (y_1^2 - 1) + |1 - y_2^2| - |(y_1 + y_2)(y_1 + y_2 - 2)|\\ &= (y_1^2 - 1) + |1 - y_2^2| - (y_1 + y_2)|y_1 + y_2 - 2|, \tag{2} \end{align*} where the last identity is due to $y_1 = |y_1| \geqslant |y_2| = -y_2$, i.e. $y_1 + y_2 \geqslant 0$.
Case 1: $y_1 + y_2 \geqslant 2$. Then by $y_1 > 1$ and $y_2 < 0$, there is\begin{align*} (2) &= (y_1^2 - 1) + |1 - y_2^2| - (y_1 + y_2)(y_1 + y_2 - 2)\\ &\geqslant (y_1^2 - 1) + (y_2^2 - 1) - (y_1 + y_2)(y_1 + y_2 - 2)\\ &= -2(y_1 - 1)(y_2 - 1) > 0. \end{align*}
Case 2: $0 < y_1 + y_2 < 2$. Then\begin{align*} (2) &= (y_1^2 - 1) + |1 - y_2^2| + (y_1 + y_2)(y_1 + y_2 - 2)\\ &\geqslant (y_1^2 - 1) + (1 - y_2^2) + (y_1 + y_2)(y_1 + y_2 - 2)\\ &= 2(y_1 - 1)(y_1 + y_2) > 0. \end{align*}
Case 3: $y_1 + y_2 = 0$. Then by $y_2 = -y_1 < -1$, there is$$ (2) = (y_1^2 - 1) + (y_2^2 - 1) - 0 = 2(y_1^2 - 1) > 0. $$
Threefore,$$ \sum_{k = 1}^{2m + 1} |1 - y_k'^2| < \sum_{k = 1}^{2m + 1} |1 - y_k^2|. $$
This implies$$ \min_{\substack{y_1, \cdots, y_{2m + 1} \in \mathbb{R}\\\sum y_k = 0}} \sum_{k = 1}^{2m + 1} |1 - y_k^2| = \min_{\substack{y_1, \cdots, y_{2m + 1} \in \mathbb{R}\\\sum y_k = 0\\|y_k| \leqslant 1\ (1 \leqslant k \leqslant 2m + 1)}} \sum_{k = 1}^{2m + 1} |1 - y_k^2|. $$
Now it suffices to prove that for any $y_1, \cdots, y_{2m + 1} \in \mathbb{R}$, if $|y_k| \leqslant 1$ for each $k$ and $\sum\limits_{k = 1}^{2m + 1} y_k = 0$, then$$ \sum_{k = 1}^{2m + 1} (1 - y_k^2) \geqslant 1, $$ i.e. $\sum y_k^2 \leqslant 2m$, which is true by the lemma.
A simpler proof of lemma: Without loss of generality, suppose $a_1, \cdots, a_k \geqslant 0$ and $a_{k + 1}, \cdots, a_{2m + 1} \leqslant 0$.
If $\sum\limits_{j = 1}^k a_j^2 \leqslant k - 1$, then$$ \sum_{j = 1}^{2m + 1} a_j^2 = \sum_{j = 1}^k a_j^2 + \sum_{j = k + 1}^{2m + 1} a_j^2 \leqslant (k - 1) + (2m - k + 1) = 2m. $$ If $\sum\limits_{j = k + 1}^{2m + 1} a_j^2 \leqslant 2m - k$, analogously there is $\sum\limits_{j = 1}^{2m + 1} a_j^2 \leqslant 2m$.
Now suppose$$ \sum_{j = 1}^k a_j^2 > k - 1, \quad \sum_{j = k + 1}^{2m + 1} a_j^2 > 2m - k. \tag{3} $$ Because$$ k - 1 < \sum_{j = 1}^k a_j^2 \leqslant \sum_{j = 1}^k a_j = -\left( \sum_{j = k + 1}^{2m + 1} a_j \right) \leqslant 2m - k, $$ then $2k < 2m + 1$, which implies $k \leqslant m$. Thus$$ m \leqslant 2m - k < \sum_{j = k + 1}^{2m + 1} a_j^2 \leqslant -\left( \sum_{j = k + 1}^{2m + 1} a_j \right) = \sum_{j = 1}^k a_j \leqslant k \leqslant m, $$ a contradiction. Therefore, (3) cannot hold.