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.
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.