Derivative of the l2-norm of a multivariate complex matrix I have a 4x1 vector with complex elements $\mathbf{c}=\left[c_{1}, \dots, c_{N}\right]^{ T }$  and the following $\ell2$-norm function  $f(\mathbf{c}) = \|\textbf{A}.\textbf{c}\|_{2}^2 = \sum_{m=1}^{M}\left|\mathbf{a}_{m}^{H}\mathbf{c}\right|^{2}=\sum_{m=1}^{M}\mathbf{c}^{H}\mathbf{a}_{m}\mathbf{a}_{m}^{H}\mathbf{c}$ , where the elements of c are complex and have the form $c_i = r_i\cdot exp(j\Phi_i)$
I am trying to find the derivatives the aforementioned function given $r_i$ and $\Phi_i$
I tried this by decomposing my complex numbers and sums but it takes too long and I cannot shorten my formulas at the end. I think there is an alternative way as this post states but I am unable to find it. Any help or hints are much appreciated. Thank you in advance.
 A: Given the real vectors $(r,\phi)$, define the complex vectors $(p,c,b)$ as
$$\eqalign{
p &= \exp(j\phi) &\implies dp = j\,p\odot d\phi \cr
c &= r\odot p    &\implies dc = r\odot dp + p\odot dr \cr
b &= A^HAc \cr
}$$
where $\odot$ denotes the elementwise/Hadamard product.
Further, a colon will denote the trace/Frobenius product, i.e. 
$\,\,A:B = {\rm Tr}(A^TB)$
(NB: The exp() function is applied elementwise)
Calculate the differential of the real function $(f)$ in terms of the real variables $(r,\phi)$. 
$$\eqalign{
 f &= (Ac)^* : Ac \cr
df
 &= (Ac)^* : (A \, dc) + (Ac) : (A \, dc)^* \cr
 &= b^*:dc + b:dc^* \cr
 &= b^*:(r\odot dp + p\odot dr) + b:(r\odot dp + p\odot dr)^* \cr
 &= (r\odot b^*):dp + (r^*\odot b):dp^* + (p\odot b^*):dr + (p^*\odot b):dr^* \cr
 &= (r\odot b^*):(j\,p\odot d\phi) + (r^*\odot b):(j\,p\odot d\phi)^*
  + (p\odot b^*):dr + (p^*\odot b):dr^* \cr
 &= (j\,p)\odot(r\odot b^*):d\phi + (j\,p)^*\odot(r^*\odot b):d\phi^*
  + (p\odot b^*):dr + (p^*\odot b):dr^* \cr
 &= 2\,{\mathcal Re}(j\,p\odot r\odot b^*):d\phi + 2\,{\mathcal Re}(p\odot b^*):dr \cr
}$$
The fact that $(dr^*=dr,\,\,d\phi^*=d\phi)$ and ${\mathcal Re(z)=\tfrac{1}{2}(z+z^*)}$ allows terms to be combined in that last line.
In this form, the gradients can be identified as
$$\eqalign{
\frac{\partial f}{\partial\phi}
 &= 2\,{\mathcal Re}(j\,p\odot r\odot b^*) \cr
 &= 2r\odot {\mathcal Im}(p^*\odot b) \cr
\frac{\partial f}{\partial r} &= 2\,{\mathcal Re}(p^*\odot b) \cr
}$$
