Prove subset of $\mathbb{C}^n$ is convex and complete I have to prove that the subset $M=\sum_{i=1}^n x_i=1$ of $\mathbb{C}^n$ is convex and complete w.r.t. the inner product $<x,y>=\sum_{i=1}^n x_i\bar{y_i}$.
Now being convex is trivial. However for completeness I want to be sure about the correctness of my proof.
My attempt:
take an $x\in \bar{M}$, and a sequence $(x_m)\subset M$ that converges to $x$. Then we know that for all $\epsilon>0$ and large enough $m$: 
$$||x_m-x||=(\sum_{i=1}^n |x_{m_i}-x_i|^2)^{\frac{1}{2}}<\epsilon$$
So for every $1\leq i\leq n$
$$|x_{m_i}-x_i|<\epsilon$$
So then we know that
$$\sum_{i=1}^n \Im{(x_{m_i})} -\epsilon n<\sum_{i=1}^n \Im{(x_{i})}<\sum_{i=1}^n \Im{(x_{m_i})} + n\epsilon$$
And since $x_m\in M$ we know $\sum_{i=1}^n \Im{(x_{m_i})}=0$ so that
$$-n\epsilon <\sum_{i=1}^n \Im{(x_{i})}<n\epsilon$$
So that $\sum_{i=1}^n \Im{(x_{i})}=0$. Now we can do the same thing to establish $\sum_{i=1}^n \Re{(x_{i})}=1$, so that $x\in M$, which proves $M$ is closed in $\mathbb{C}^n$, and thus $M$ is complete.
If someone could tell me if this proof is correct that would be great. Thanks

Let $\varphi : x = (x_i)_{1\leq i \leq n} \mapsto \sum_{i=1}^n x_i$. This map is continuous and $M = \varphi^{-1}( \{1\})$. As $\{1\}$ is closed, $M$ is also closed, being inverse image of a closed set by a continuous map. And as you said, it therefore complete.
