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Let $f(z)=\begin{cases} \frac{z^5}{\left | z \right |^4} & \text{ if } z\neq 0 \\ 0 & \text{ if } z=0 \end{cases} $

I could show this is continuous on $\mathbb{C}$.

And, I would like to show this satisfies the C-R equation at $z=0$.

However, there are some things confusing me, especially the meaning of 'a function satisfies C-R equation at some specific point.'

Let me share my work.

Since $f$ is continuous on $\mathbb{C}$, we can let $f(z) = u(x,y) + iv(x,y)$ for some continuous $u,v$ on $\mathbb{C}$.

Since $f(0) = 0, \lim_{x\rightarrow 0, y\rightarrow 0}u(x,y)=0$ and $\lim_{x\rightarrow 0, y\rightarrow 0}v(x,y)=0$.

This means $u(0,0) = 0$ and $v(0,0)=0$.

Thus, $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y} =0$

Hence, the C-R equation satisfies for $f$ at $z=0$.

Is this how I should approach for this kind of problems?

Or should I let $z = x+yi$ and do calculations? I also tried this way, but it is not that simple as I expected. This is why I want to see if this proof works.

john
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    $u(0) = v(0) = 0$ does not imply that $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y} =0$ – Martin R Sep 23 '21 at 06:43
  • @MartinR Then, I think I should use the definition "$\frac{\partial f}{\partial x}(z_0)=\lim_{h\rightarrow 0}\frac{f(z_0+h)-f(z_0)}{h}$, am I right? With $f$ replaced by $u$ or $v$? – john Sep 23 '21 at 06:47
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    Yes, exactly. $\frac{\partial u}{\partial x}(0)=\lim_{h\to 0, h \in \Bbb R}\frac{u(h)-u(0)}{h}$, and similar for the other derivatives. – Martin R Sep 23 '21 at 06:51
  • @MartinR Thanks for the clarification! – john Sep 23 '21 at 06:52
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    See https://math.stackexchange.com/a/2924208/42969 for a correct solution. – Martin R Sep 23 '21 at 07:04

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