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If we have a region R is $f(z)$ analytic in the region R if and only if it satisfy the Cauchy-Riemann equations for every point in R. If not what are the other conditions it must satisfy? Do we have to check that $f'(z)$ is continuous or is that given by the Cauchy-Riemann equations?

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The standard result is that for functions $f$ of class $C^{1}$, $f$ is complex analytics iff $u,v$ satisfy the CR equations. There is however, a more powerful result: The Looman-Menchoff theorem allows us to replace $C^1$ above with $C^0=C$ (=the class of continuous functions).

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  • $\begingroup$ +1, don't we need to assume that $f$ is essentially differentiable though--else how are we plugging $f$ into the Cauchy-Riemann equations--unless this is meant in some distributional sense. $\endgroup$ – Alex Youcis Nov 30 '14 at 8:27
  • $\begingroup$ So we are saying that the standard results states that if $f$ satisfies the CR and $f'$ is continuous in the region R then $f$ is analytic in that region. But in fact from the Looman-Menchoff theorem we can state that $f$ is analytic in the region R if $f$ is continuous in the region R and satisfies the CR equations (and this is a necessary and sufficient condition?). Is this correct? $\endgroup$ – user135842 Nov 30 '14 at 8:27
  • $\begingroup$ @AlexYoucis The existence of partial derivatives does not imply that a function is differentiable. You do need to assume the partial derivatives exist (except perhaps on a countable set). $\endgroup$ – Robert Israel Nov 30 '14 at 8:36
  • $\begingroup$ @RobertIsrael Thank you, I was being imprecise. I didn't meant that the total derivative exists, I meant that the individual partials exist. Regardless, whether we assume they exist off a countable set, or otherwise, it seems like literally continuity doesn't make sense. I suppose that 'and satisfies the Cauchy-Riemann equations' should mean that the partial derivatives exist, as well. I'm being slightly pedantic, but I just want to make sure I understand. $\endgroup$ – Alex Youcis Nov 30 '14 at 8:39

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