How to minimize $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 - z_4|^2$? 
If $z_1,z_2,z_3,z_4 \in \mathbb{C}$ satisfy $z_1 + z_2 + z_3 + z_4  =
 0$ and $|z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 = 1$, then the least value of $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 -
 z_4|^2$ is $2$.

I solved this by taking two of the values to be $\frac{1}{2}$ and other two as $-\frac{1}{2}$. But I couldn't prove this in an efficient manner.
 A: Let $\vec w$ be related to $\vec z$ by the self-inverse Hadamard transform:
$$H = \frac{1}{\sqrt4}
\begin{pmatrix}
1&1&1&1\\
1&1&-1&-1\\
1&-1&1&-1\\
1&-1&-1&1
\end{pmatrix}$$
Then $w_1 = 0$, i.e. we have reduced the problem to three dimensions. $H$ is unitary so $|\vec w|^2 = |\vec z|^2 = 1$.
Let us now look at the terms of the objective function.
$$z_1-z_2 = w_3+w_4$$
$$z_1-z_4 = w_2+w_3$$
$$z_2-z_3 = w_2-w_3$$
$$z_3-z_4 = w_3-w_4$$
Then:
$$|z_1-z_2|^2+|z_1-z_4|^2 = 2(|w_3|^2+|w_4|^2)$$
$$|z_2-z_3|^2+|z_3-z_4|^2 = 2(|w_3|^2+|w_2|^2)$$
So the objective function is equal to $2+2|w_3|^2$.
A: Definitely not more efficient, but here goes anyway:
Let $z \in \mathbb{C}^4$, $e=(1,...,1)^T$, $A= \begin{bmatrix} 1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ -1 & 0 & 0 & 1 \end{bmatrix}$.
Then the problem is $\min \{  \|Az\|^2  \ | \  e^T z = 0, \|z\|^2 = 1 \}$. Some tedious multiplication shows that $A^* A = A A^*$, so $A$ is unitarily diagonalizable, and $\ker A = \operatorname{sp} \{ e \}$.
Since $A$ is unitarily diagonalizable, we can write $A z = \sum_{k=1}^3 \lambda_k (u_k^* z ) u_k$, where the $\lambda_k$ are non-zero, the $u_k$ are orthonormal, and $u_k^* e = 0$ for all $k$. Furthermore, $\|A z\|^2 = \sum_{k=1}^3 |\lambda_k|^2 |u_k^* z|^2$. If we let $\sigma = \min_k |\lambda_k|^2$, we have
$\|A z\|^2 \ge \sigma\sum_{k=1}^3  |u_k^* z|^2$. If $e^T z = 0$, we have $z \in \operatorname{sp} \{ u_1, u_2, u_3\}$, and so $\sum_{k=1}^3  |u_k^* z|^2 = \|z\|^2$.
Hence if $e^T z = 0$, we have $\|A z\|^2 \ge \sigma \|z\|^2$, and if $u$ is the eigenvector that corresponds to $\sigma$, then we have $\|Au\|^2 = \sigma \|u\|^2$.
Hence the minimum value is the square of the modulus of the smallest non-zero eigenvalue of $A$, and the corresponding unit eigenvector is a minimizer.
We have $\chi_A(s) = x(x-2)(x^2-2x+2)$, hence the non-zero eigenvalues are $2, 1\pm i$, and so we have $\sigma = 2 = |1\pm i|^2$. The corresponding eigenvector is $(i,-1,-i,1)^T$, hence a minimizer is $z = \frac{1}{2}(i,-1,-i,1)^T$.
A: Derive respect $z_{i}^{*}$ with the conditions $\sum_{i}z_{i}^{*} = 0$ and
$\sum_{i}z_{i}^{*}z_{i} - 1 = 0$ by using Lagrange multipliers. In this way,we get the values of $\left\lbrace z_{i}\right\rbrace$. Plug these values in the expression to get the minimum.
A: I guess writing in terms of vectors makes it easier. Let $A,B,C,D$ be points in $\mathbb C$ with affixes $z_1,z_2,z_3,z_4$. Write $a=\vec{OA}, b=\vec{OB}, c=\vec{OC}, d=\vec{OD}$.
We then have $\displaystyle\sum_{cyc}a\cdot a=1$ and $\displaystyle\sum_{cyc}a=0$. Surprisingly,
$$\begin{aligned}\sum_{cyc}|z_1-z_2|^2&=\sum_{cyc}(\vec{OA}-\vec{OB})\cdot(\vec{OA}-\vec{OB})=\sum_{cyc}(a\cdot a + b\cdot b - 2a\cdot b).\\&=2-2(a\cdot b + b\cdot c + c\cdot d + d\cdot a)\\&=2-2\Bigl\{a\cdot b + b\cdot c + c\cdot (-a-b-c)+ a\cdot (-a-b-c)\Bigr\}\\& = 2+2(a\cdot a+2a\cdot c + c\cdot c)\\& = 2+2(a+c)\cdot(a+c)=2+2|z_1+z_3|^2.\end{aligned}$$
This is minimum when $z_1=-z_3, z_2=-z_4, |z_1|^2+|z_2|^2=\frac 12. \ \ \Box$
