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I've been trying to understand the complex function derivatives, but there are some issues that I'm encountering. I'm following the Ahlfors Complex Analysis. When he is demonstrating the derivative via the limit, he does it both with a pure infinitesimal Real and a pure Imaginary value, but I don't get why should it be concluded that they must have the same value since you are going in different directions of the complex plane (with the real value you are going in the real axis and with the imaginary value you are going with the imaginary axis).

Also, I see some similarities with the multivariate calculus, without being the same, and that might be something that is causing me to some misconception.

pdaranda661
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    if your assumption is that the complex derivative already exists at some $z=\zeta$, then you can travel any path converging to that $\zeta$ with your difference quotient, it will be the same $f'(\zeta)$, in the limit. But you are right to be suspicious, for the two paths along the axes are not sufficient. – 311411 Jun 01 '21 at 21:35
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    The derivative of a function $f$ at at point $z$ is defined as the limit, if it exists of $$f'(z)=\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$$If the limit exists, then it must be the same whether $\Delta z=\Delta x$ or $\Delta z=i\Delta y$, where $\Delta x$ and $\Delta y$ are real. – Mark Viola Jun 01 '21 at 21:37
  • @MarkViola but that is something I don't totally get. Why it has to be strictly the same? isn't an imaginary term telling us how it varies in the imaginary axis, and a real term how it changes in the real axis? – pdaranda661 Jun 04 '21 at 06:12
  • @311411 is there a demonstration to show that is true for any direction on the axes? – pdaranda661 Jun 04 '21 at 06:13
  • @pdaranda661 If the limits along real and imaginary axes are different, then the function $f'(z)$ fails to exist. That is to say, $f$ is NOT differentiable at $z$. If it is differentiable, by definition the limit along all directions must be equal. If not, the function is NOT differentiable. – Mark Viola Jun 04 '21 at 16:01

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There definitely are some connections between multi-variable calculus and complex analysis, and your multi-variable experience will help you to build intuition, particularly in the first few weeks of the class.

You are right, we could talk about functions of complex variables, thinking about the real part as being independent from the complex part, and we have partial derivatives with respect to each part.

But, there is nothing that keeps us from creating a more restricted definition of the derivative. And there is a class of functions where this limit exists:

$f'(z) = \lim_\limits{|\Delta z| \to 0} \frac {f(z+\Delta z) - f(z)}{\Delta z}$

Some examples are polynomial functions of a complex variable and the exponential function. Complex analysis studies the behavior of these functions.

Doug M
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  • Okay, let's see if I got it. If you did the derivative of a complex function taking the denominator as a real number and another time as an imaginary number, you would obtain how it varies in each one of the axes, but, if you want them to be considered the Complex Derivative, then those have to be the same, no? It seems a bit weird that the derivative's definition changes to be a bit different, although it is normal that in real functions it cannot be taken into account the variation with an imaginary term. I don't know, I think I'm taking some things for granted that don't apply here – pdaranda661 Jun 04 '21 at 06:08