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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 $\Delta re^{i\Delta\theta}$?

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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)

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  • $\begingroup$ Thanks! How do I write $\Delta x$ and $\Delta y$ in terms of $\Delta r$ and $\Delta \theta$? $\endgroup$ Aug 21 '18 at 19:17
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    $\begingroup$ Since $x(r,\theta)=r\cos\theta$ and $y(r,\theta)=ri\sin\theta$ you can differentiate $x,y$ in terms of $r,\theta$. For example: $\Delta x=\frac{\partial x}{\partial r}\Delta r+ \frac{\partial x}{\partial\theta}\Delta\theta$ $\endgroup$ Aug 21 '18 at 19:26

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