Given the complex-valued function $$\begin{align*}f: \mathbb C&\to \mathbb C\\z&\mapsto z\mathrm{Re}(z),\end{align*}$$ I want to find the points at which $f$ is differentiable using only the difference quotient, and compute the derivative at those points.
The derivative is $$f'(z)=\lim\limits_{h\to0}\frac{(z+h)\mathrm{Re}(z+h)-z\mathrm{Re}(z)}{h},$$ but how should I continue from here? More specifically, what does it mean for a complex-valued number like $h$ to tend to $0$ (unlike on the real axis, there are many ways to approach $0$) and how do I handle this when calculating a limit?
I have seen a trick using polar coordinates and simply letting $r$, a real number, tend to $0$, but I think this can only be used if I already know the point at which I want to calculate the derivative.
\begin{align} f'(z) &=\lim\limits_{h\to0}\frac{(z+h){\bf Re}(z+h)-z{\bf Re}(z)}{h}\\ &=\dfrac12\lim_{h\to0}\frac{h(2z+h)+h\bar{z}+z\bar{h}+h\bar{h}}{h}\\ &=\dfrac12\lim_{h\to0}\left(2z+h+\bar{z}+\bar{h}+z\frac{\bar{h}}{h}\right)\\ &=z+\dfrac12\bar{z}+\dfrac12z\lim_{h\to0}\frac{\bar{h}}{h} \end{align} this limit exists only when $z=0$.