Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.