Inverse of complex vector How is the inverse of a complex vector calculated?
In $\Bbb R$, the inverse vector
$
        X=
        \begin{bmatrix}
        x_1 \\
        x_2 \\
        \vdots \\
        x_n \\
        \end{bmatrix}
$
is calculated like this:
$
l=\sum_1^n{x_i^2} \\
X_{inv}=
        \begin{bmatrix}
        \frac{x_1}{l} \\
        \frac{x_2}{l}\\
        \vdots \\
        \frac{x_n}{l} \\
        \end{bmatrix}
$
Does this rule apply also on complex vectors (any $x_i$ are complex)?
EDIT:
A real number's inverse is $\frac{1}{number}$, and matrices has also inverses. Therefore it's defined for "dimensions" 0 and 2.
So I'm asking myself what are the rules for "dimension" 1 (vectors)?
Background: I'm implementing a generic variable system (C#) that can handle numbers, vectors and matrices, both real and complex. I want to implement all operators that are possible. This question came up when implementing the inverse operation, but I know it only for numbers and matrices.
 A: You are looking for an unit vector, so a vector $v$ with $ ||v|| = 1 $. 
This concept can be extended to any normed vector space, by setting $$x_{normed} = \frac 1 { ||x||} x $$ Then $||x_{normed}|| = 1$.
In your example you have $||z|| = \sum\limits_{i=1}^n |z_i|^2$
As for inverses in the case of "multiplicative inverse", there is (usually) none for vecotors. You would need a function $f: \mathbb R ^n \times \mathbb R ^n \to \mathbb R^n$, which you don't have. 
A: One way to consider this is regarding vector $v$ as $n$ by 1 matrix. For matrix, we have Moore-Penrose pseudoinverse for it. That is,
$$v^{\dagger}=(v^*v)^{-1}v^*=\frac{v^*}{||v||^2}$$
Another way is to consider in Clifford algebra $\mathcal{C}\ell(V)$, in which a quadratic form $Q(x)=x^2\in K$ is defined for every $x\in V$, where $V$ is a vector space endowed with scalar field $K$. Then vector $v$ admits, if there exists, a unique inverse $u$ which can be written as
$$u=\frac{v}{Q(v)}$$
The example you mentioned is actually taking $Q(v)=||v||^2$. This can be extended to complex cases.
A: Let $X=(x_1,\dots,x_n)$ be a vector in $\mathbb C^n$. We search for a complex vector $Y=(y_1,\dots,y_n)$ s.t. 
$$\langle X,Y\rangle=1 $$
where we introduced the sesquilinear form $\langle X,Y\rangle=\sum_{i=1}^n \bar{x}_iy_i$.
Choosing
$$y_i= \frac{x_i}{||X||^2},$$
with $||X||^2=\langle X, X\rangle$ leads to the thesis.
