The gradient of $\| x - \alpha \nabla f(x) \|_2^2$ with respect to $x$, where $x \in \mathbb{C}^n$, $\nabla f(x) \in \mathbb{C}^n$, and $\alpha \in \mathbb{R}$?
Following greg's comments, please see below my attempt. Is it correct?
Let us say that $A : B := {\mathrm{Tr}}(A^{\rm T} B)$, $(\cdot)^*$ means complex conjugate, $g := \nabla f$, $H := \nabla g$.
And, $\phi := \| x - \alpha \nabla f(x) \|_2^2 = \| x - \alpha g \|_2^2 = \left(x - \alpha g \right) : \left(x - \alpha g \right)^*$.
Then, ignoring the non-conjugate term in the differential, \begin{align} d \phi &= \left(x - \alpha g \right) : d\left(x - \alpha g \right)^*\\ &= \left(x - \alpha g \right) : \left(dx^* - \alpha \, H^* dx^* \right) \\ &= \left(I - \alpha H^* \right)^{\rm T}\left(x - \alpha g \right) : dx^* \end{align}
So, the gradient is $$\frac{\partial \phi}{\partial x^*} = \left(I - \alpha H^* \right)^{\rm T}\left(x - \alpha g \right) = \left(I - \alpha H^{\rm H} \right)\left(x - \alpha \nabla f \right)$$
Is this correct?