Complex derivative of Hadamard product inside Frobenius norm I'm trying to find the complex derivative of 
$$||R - P \circ \gamma \gamma ^H||_F ^2$$.
with respect to $\gamma$. I saw the post regarding the real counterpart of the same question here. However, when I tried applying similar principles given there $-2(P\circ M+P^T\circ M^T)\gamma$,  (where $M=R - P \circ \gamma \gamma^T$), it didn't work. Here $||.||_F$ is the Frobenius norm and $\circ$ is the Hadamard product.
 A: As in the linked answer, define the matrix
$$\eqalign{
M   &= P\circ\gamma\gamma^H &-\; R \\
}$$
Various conjugations will also prove useful
$$\eqalign{
M^H &= P^H\circ\gamma\gamma^H &-\; R^H \\
M^T &= P^T\circ\gamma^*\gamma^T &-\; R^T \\
M^* &= P^*\circ\gamma^*\gamma^T &-\; R^* \\
}$$
where $M^*$ denotes the complex conjugate.
Calculate the gradient of the function wrt $\gamma$ while pretending that $\gamma^H$ is constant. 
$$\eqalign{
 f &= \|M\|^2_F = M^*:M \\
df
 &= M^*:dM + M:dM^* \\
 &= M^*:dM + M^T:dM^H \\
 &= M^*:(P\circ d\gamma\gamma^H) + M^T:(P^H\circ d\gamma\gamma^H) \\
 &= (P\circ M^*)\gamma^*:d\gamma + (P^H\circ M^T)\gamma^*:d\gamma \\
\frac{\partial f}{\partial\gamma}
 &= (P\circ M^*+P^H\circ M^T)\gamma^* \\
}$$
Since $f$ is real, we immediately know that
$$\eqalign{
\frac{\partial f}{\partial\gamma^*}
 &= \left(\frac{\partial f}{\partial\gamma}\right)^*
 &= (P^*\circ M+P^T\circ M^H)\gamma \\
}$$
and 
$$\eqalign{
\frac{\partial f}{\partial\gamma^H}
 &= \left(\frac{\partial f}{\partial\gamma^*}\right)^T
 &= \gamma^T(P^H\circ M^T+P\circ M^*) \\
\\
}$$
NB:$\;$ Changes in $\left(\gamma, \gamma^*, \gamma^H\right)\,$ 
are perfectly correlated, so the total differential is 
$$\eqalign{
df
 &=    \left(\frac{\partial f}{\partial\gamma}\right)  :d\gamma
 \;+\; \left(\frac{\partial f}{\partial\gamma^*}\right):d\gamma^* \\
}$$
In the linked answer, $\gamma=\gamma^*\,$ so both terms are identical and can be combined.
