How do I find the derivative of the $l1$-norm of a vector of complex numbers with respect to the vector? I want to take the derivative complex vector $x^*$ with respect to the vector $x\cdot \sqrt{x^{*}x}$. If anyone can tell me that will be great.
 A: Unfortunately, the one norm of a complex vector (the sum of the absolute values of the entries of the vector) is not a differentiable function.  In fact, the absolute value of a scalar complex number z=x+i*y is not
a differentiable function.  
To see this, use the Cauchy-Riemann conditions.  Write the absolute value as 
abs(z)=abs(x+i*y)=sqrt(x^2+y^2)+i*0
let u(x,y)=sqrt(x^2+y^2) (the real part) and v(x,y)=0 (the imaginary part.)
If the absolute value was differentiable then it would satisfy the Cauchy-Riemann conditions.  In particular, the partial derivative of u with respect to x would have to equal the partial derivative of v with respect to y.  Since this clearly doesn't hold, there's no need to check the other half of the CR conditions, and you can conclude that abs(z) is not differentiable.  
A: If $f_k(x) = |x_k|$, then
${\partial f_k (x) } = \begin{cases}
\{-e_k\}, & x_k < 0 \\
[-e_k,e_k], & x_k = 0 \\
\{e_k\}, & x_k > 0 \end{cases}$.
If $f(x) = \|x\|_1 = \sum_k f_k(x)$, then
$\partial f(x) = \sum \partial f_k(x)$.
The subgradient is a singleton at $x$ iff $f$ is differentiable at $x$ 
iff $x_k \neq 0$ for all $k$.
