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.

Is following contradictory?

"Then $f = u + iv$ is complex-differentiable at that point if and only if the partial derivatives of $u$ and $v$ satisfy the Cauchy–Riemann equations (1a) and (1b) at that point. The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point." - 1[1, second paragraph]

They say that $f$ is complex-differentiable iff partial derivatives of $u$ and $v$ satisfy C-R equations, but still it is not enought to ensure complex differentiability at that point.

So do you need extra conditions as wikipedia says or not for $f$ to be complex differentiability? Can you give me example, where function satisfy C-R equations, but is not Complex differentiable at certain point?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

For an explicit example, let $$ f(z) = \begin{cases} \exp(-1/z^4) & z \neq 0 \\ 0 & z = 0\end{cases} $$

You can check that $f$ satisfies Cauchy-Riemann's equations everywhere, but $f$ is not real-differentiable (or even continuous) at $z=0$.

You may also be interested in Looman-Menchoff's theorem which shows that it's enough to assume that $f$ is continuous and satisfies Cauchy-Riemanns equations everywhere to conclude that $f$ is holomorphic. (Note that in Looman-Menchoff, it's not enough to assume continuity and CR at a point.)

share|improve this answer

You left out the part where it says, "Suppose that $u$ and $v$ are real-differentiable...." It says that in the absence of that assumption, the existence of partials satisfying C-R is not enough to insure complex differentiability. So to find the example you want, you'll first have to find functions with partials but not real-differentiable.

share|improve this answer
    
Maybe I'm missing something, but how can they not be real differentiable if they satisfy C-R? –  Thomas Andrews Jun 14 '13 at 12:30
    
You are missing the possibility that a function $u(x,y)$ can be such that its partials with respect to $x$ and to $y$ both exist at some point, but it isn't differentiable at that point. Review the definition of differentiability for multivariate functions. –  Gerry Myerson Jun 14 '13 at 12:37
    
Yes. So to find that eq. |x| is not real-differentiable at $x=0$, so I think you can't find complex function of whose u and v would not be real-differentiable? –  laovultai Jun 14 '13 at 12:44

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.