I've been working through this set of notes on differentials of $\mathbb{R}$-valued functions of complex variables (and this MSE question), and I'm trying to work through a simple example. Do I have the following terminology and computations right, or have I messed up something?
Let $A \in \mathbb{R}^{n \times n}$ with $A = A^{\top}$. Define $z := r e^{j \theta}$ and $f(z) := \frac{1}{2} z^H A z$. For the differential of $f$ with respect to $x$ for $x = (r,\theta)$, we have operators $$ \frac{\partial}{\partial r}(\cdot) = \frac{\partial}{\partial z}\frac{\partial z}{\partial r}(\cdot) + \frac{\partial}{\partial \bar{z}}\frac{\partial \bar{z}}{\partial r}(\cdot) = e^{j\theta} \odot \frac{\partial}{\partial z}(\cdot) + e^{-j\theta} \odot \frac{\partial}{\partial \bar{z}}(\cdot) \tag{1} $$ and $$ \frac{\partial}{\partial \theta}(\cdot) = \frac{\partial}{\partial z}\frac{\partial z}{\partial \theta}(\cdot) + \frac{\partial}{\partial \bar{z}}\frac{\partial \bar{z}}{\partial \theta}(\cdot) = j r e^{j\theta} \odot \frac{\partial}{\partial z}(\cdot) - j r e^{-j\theta} \odot \frac{\partial}{\partial \bar{z}}(\cdot) \tag{2} $$ so that $$ \frac{\partial}{\partial r} f = e^{j\theta} \odot \frac{1}{2} z^H A + \frac{1}{2} e^{-j\theta} \odot Az \tag{3} $$ $$ \frac{\partial}{\partial r} f = \frac{1}{2} j z \odot \frac{1}{2} z^H A + \frac{1}{2} j z \odot Az \tag{4}. $$
Is it correct to say that the operators in (1) and (2) are nonlinear because $\frac{\partial}{\partial z}(\cdot)$ and $\frac{\partial}{\partial \bar{z}}(\cdot)$ are nonlinear (see eq. 20 and 51 here)?
Is the $\mathbb{C}\mathbb{R}$-Hessian w.r.t. $r$ and $\theta$ found by just applying (1) and (2) to the parts of (3) and (4) involving $z$, and normal $\mathbb{R}$-calculus to the parts involving $r$ and $\theta$?