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What's the intuition behind $f'(z)= \frac{\partial f}{\partial x}$ and $f'(z)= \frac{1}{i} \frac{\partial f}{\partial y}$, where $f : \mathbb{C} \to \mathbb{C}$ is assumed to be analytic?

I know the algebraic proof of this, but I am not sure what these equations really mean. If anyone could explain why these equations should be true, I would be grateful!

EDIT : Having thought about it a bit more I think these equations are not so dramatic. For an analytic function the derivative must agree with the directional derivatives ( this can probably be said better ) and this is, I suppose, just what these equations are saying.

  • No, I seek understanding of the seperate equations, not of the proof of these equations or some relationship between them. –  Dec 19 '19 at 13:51
  • Ok, that makes sense. –  Dec 19 '19 at 13:53
  • I am a bit confused, why do you write a partial derivative w.r.t $z$? $f$ only has one variable.. – Strichcoder Dec 19 '19 at 13:54
  • @Yourong Zang Follows more or less from this post https://math.stackexchange.com/questions/314863/what-is-the-intuition-behind-the-wirtinger-derivatives?fbclid=IwAR2j-g7gJZMqm5yACzhDAiOpeSuF7oydhbfdIFPD2ZVSv6U1f8eXomc1izQ –  Dec 19 '19 at 13:54
  • @ Strichcoder Sorry will edit it. No reason for this to be partial! –  Dec 19 '19 at 13:54
  • @Strichcoder if $f$ is analytic, i.e., complex-differentiable, $\frac{df}{dz}=\frac{\partial f}{\partial z}$. So it's ok to write a partial derivative. –  Dec 19 '19 at 13:58
  • @Strichcoder yes, but best not cause confusion! –  Dec 19 '19 at 13:58
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    Made an edit to my original post! –  Dec 19 '19 at 14:10

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$f'(z) = \lim_{h \to 0} {f(z+h)-f(z) \over h }$, where $h \in \mathbb{C}$.

Hence $f'(z) = \lim_{t \to 0, t \in \mathbb{R}} {f(z+t)-f(z) \over t } = \partial_x f(z)$ and $f'(z) = \lim_{t \to 0, t \in \mathbb{R}} {f(z+it)-f(z) \over it } = {1 \over i}\partial_y f(z)$.

copper.hat
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