I want to find $\nabla_x f(x)$ where $f(x) = f(z(x)) = z(x)^H\, Y\, z(x)$ and $z(x) = c \exp(\text{i}\, x)$ for $c \in \mathbb{R}^n$, $x \in \mathbb{R}^n$, and $z^H := (z^*)^T := (\bar{z})^T$. I tried to use the "normal" chain rule as $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial z} \frac{\partial z}{\partial x} = Y^* z^* \cdot \text{diag}(\text{i}\, z) $$ from $$ \begin{aligned} \frac{\partial f}{\partial x_k} &= \frac{\partial f}{\partial z_k} \frac{\partial z_k}{\partial x_k} \\ &= \sum_i \bar{z}_i \, \sum_j \frac{\partial}{\partial z_k} \frac{\partial z_k}{\partial x_k} z_j y_{ij} \\ &= \sum_i \bar{z}_i \, \sum_j \frac{\partial z_k}{\partial x_k} \delta_{jk} y_{ij} \\ &= \sum_i \bar{z}_i \,\frac{\partial z_k}{\partial x_k} y_{ik} \\ &= \sum_i \bar{y}_{ki} \, \bar{z}_i \,\text{i}\, c_k \exp(\text{i}\, \, x_k) \\ &\implies \frac{\partial f}{\partial x} = Y^* z^*\, \text{diag}(\text{i}\, c \, \exp(\text{i}\, x)). \end{aligned} $$ Is this correct, or did I mess something up? I based the derivation on this thesis (p100) and this answer.
If instead $Y \in \mathbb{R}^{n\times n}$ and is symmetric, should the gradient w.r.t $x$ have any imaginary component?