I have this defintion
$$\dfrac{d}{dz}f(z)=\lim\limits_{\Delta z\to0}\dfrac{f(z+\Delta z)-f(z)}{\Delta z}$$
and I know how to rewrite it in terms of the Cartesian representation of complex numbers as
$$\dfrac{d}{dz}f(z)=\lim\limits_{\substack{\Delta x\to0\\\Delta y\to0}}\dfrac{f\big((x+iy)+(\Delta x+i\Delta y)\big)-f\big((x+iy)\big)}{\Delta x+i\Delta y}$$
My question is, what is the corresponding expression for $z=re^{i\theta}$? Do I just write $\Delta z$ as $\Delta re^{i\Delta\theta}$?
In the Cartesian case $z$ was a function of two variables $z(x,y)=x+iy$, and the partial derivatives are $\frac{\partial z}{\partial x}=1,\frac{\partial z}{\partial y}=i$, that is the gradient of $z(x,y)$ is constant and equal to the vector $(1,i)$ so that applying ([1]) $dz$ to $(\Delta x,\Delta y)$ gives $\Delta x+ i \Delta y$. The same reasoning will hold in the case of polar coordinates. Here: $$z(r,\theta)=re^{i\theta}$$ so $$\frac{\partial z}{\partial \theta}=rie^{i\theta},\quad \frac{\partial z}{\partial r}=e^{i\theta}$$ hence, the result of applying $dz$ in these coordinates to $(\Delta r,\Delta\theta)$ is: $$e^{i\theta}\Delta r + rie^{i\theta}\Delta\theta$$
[1]("Applying" the gradient to a vector means taking the inner-product of the gradient with that vector)