# Limit of $u(x,y),v(x,y)$ at $0$ where $f(z)=u+iv$

I have an exercise with the following function

$$f(z)=\begin{cases} \frac{z^{5}}{|z|^{4}} & z\neq0\\ 0 & z=0 \end{cases}$$

I have prove that Cauchy-Riemann equations are satisfied at $z_{0}=0$ but that $f$ is not differentiable at $z_{0}=0$ .

I have a theorem in my notebook that claims that if $u,v$ are defined in a neighborhood of $(x_{0},y_{0})$, and $u_x,v_x$ are continuous there and satisfy C-R equations then $f=u+iv$ is differentiable at $z_{0}=(x_{0},y_{0})$.

I concluded that $u_x,v_x$ are not continuous at $(0,0)$ and I wish to prove it for the sake of practice.

The problem I am having that I can't really calculate those derivatives explicitly, I get them as limits - and taking the limit (at 0 ) of the derivative (which is written as a limit on its on) is giving me trouble

Can someone please help me out in proving that $u_x$ or $v_x$ are not continuous at $(0,0)$ ? (assuming that this conclusion I made is right)

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Isn't a differentiability criterium missing from your argument?

If $z=x+iy\to 0$ then $|f(z)|=|z|\to 0$, so it is continuous.

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I miss-copied the theorm about differentiability , I will edit the post. Thanks for pointing that out – Belgi Feb 8 '13 at 22:48
I have edited the question, thanks again – Belgi Feb 8 '13 at 22:53
Yes, but what is this $u_x,\, v_x$? The derivatives of $u$ and $v$, respectly, no? Now, I guess, these doesn't exist in $(0,0)$. – Berci Feb 9 '13 at 12:39
Yes, those are the derivatives. Since I have verified C-R equations I can tell you both does exist (and are easy to calculate) – Belgi Feb 9 '13 at 13:38
So the case is that they exist there - but not continuous there – Belgi Feb 9 '13 at 13:39