Uniqueness of a vector with the minimal Euclidean norm Let $W_{n}=\{(z_{1},\dots,z_{n})\in \mathbb{C}^{n}: z_{1}+\dots+z_{n}=1\}$. Find $z\in W_{n}$ with the minimal Euclidean norm. Is it unique? Is $W_{n}$ bounded?
If my reasoning is correct, I need to find the minimal value of $a_{1}^{2}+\dots+a_{n}^{2} $ with the condition $a_{1}+\dots+a_{n}=1$ where $a_{k}$'s are real parts of $z_{k}$'s (the sum of imaginary parts is $0$) and the answer follows from Cauchy-Schwarz, the sum of squares equals ${1}/{n}$, so an example of $z$ satisfying the conditions is $z=({1}/{n},\dots,{1}/{n})$.
But I have problems answering the other questions, could somebody give me a hint?
 A: *

*$W_n$ has a unique element $z$ with minimal Euclidean norm.
OP's arguments are right.  To make them a proof, we apply Cauchy-Schwarz on $\Bbb C$.  Suppose that $(z_1,\dots,z_n) \in W_n$.
\begin{align}
\left|\sum_{i=1}^n z_i \cdot 1 \right|^2 &\le \sum_{j=1}^n |z_j|^2 \sum_{k=1}^n 1^2 \\
\left|\sum_{i=1}^n z_i \right|^2 &\le \sum_{j=1}^n |z_j|^2 \sum_{k=1}^n 1 \\
1 &\le \sum_{j=1}^n |z_j|^2 \cdot n \\
\sum_{j=1}^n |z_i|^2 &\ge \frac1n
\end{align}
Equality holds (i.e. Euclidean norm of $(z_1,\dots,z_n)$ is minimal) iff $(z_1,\dots,z_n)=\alpha(1,\dots,1)$ for some $\alpha\in\Bbb C$.
\begin{align}
\sum_{i=1}^n z_i &= \sum_{i=1}^n \alpha \\
1 &= n\alpha \\
\alpha &= \frac1n
\end{align}
So $z=(z_1,\dots,z_n)=\frac1n(1,\dots,1)$ is the unique element in $W_n$ with minimal Euclidean norm.

*Using uniquesolution's idea, we construct an unbounded sequence $(w^{(n)}_k)_{k \in \Bbb N}$ in $W_n$ for each $n\ge2$.
$$w^{(n)}_k=(k+1,-k,\underbrace{0,\dots,0}_{n-2 \, 0's})\in W_n\quad\forall k \in \Bbb N$$
It's clear that $\lVert w^{(n)}_k\rVert_2 > \sqrt2 k \to +\infty$ as $k \to +\infty$, so for each $n \in \Bbb N$, $W_n$ contains an unbounded sequence $(w^{(n)}_k)_{k \in \Bbb N}$, so $W_n$ is unbounded unless $n=1$.

A: Let $e=(1,1,1,\cdots,1) \in \mathbb{C}^n$. Then $W_n$ may be described as
$$
            W_n=\{ z \in \mathbb{C}^n : \langle z, e \rangle = 1 \},
$$
where $\langle\cdot,\cdot\rangle$ denotes the inner product on $\mathbb{C}^n$.
It is easy to check that $\frac{1}{n}e \in W_n$. And, for any $w\in W_n$,
$$
        \langle w-\frac{1}{n}e, e \rangle = 0.
$$
Every $w\in\mathbb{C}^n$ may be written as $w=w-\frac{1}{n}e+\frac{1}{n}e$. Hence,
$$
         \|w\|^2 = \|w-\frac{1}{n}e\|^2+\|\frac{1}{n}e\|^2 \ge \|\frac{1}{n}e\|^2,
$$
and equality holds iff $w=\frac{1}{n}e$. So the unique element $w$ of minimum norm in $W_n$ is $w=\frac{1}{n}e$. That minimum norm is $\|w\|=\|\frac{1}{n}e\|=1/\sqrt{n}$. $W_n$ is not bounded because it is a vector translation of a subspace.
