1
$\begingroup$

Looman-Menchoff theorem states that a continuous complex-valued function $f(z)=u(x,y)+iv(x,y)$ defined in an open set of the complex plane is holomorphic if and only if it satisfies the Cauchy–Riemann equations.

We know that a differentiable function is continuous. So if the real and imaginary parts of the function $f(z)=u(x,y)+iv(x,y)$ are differentiable, then the complex function $f$ will be continuous, and therefore the assumption of continuity in the Looman-Menchoff theorem is unnecessary. Am I right?

$\endgroup$
  • 1
    $\begingroup$ You might appreciate this Monthly article. $\endgroup$ – user98602 Dec 27 '14 at 21:16
1
$\begingroup$

If $f$ is differentiable (viewed as a function $\mathbb{R}^2 \to \mathbb{R}^2$) then $f$ is indeed continuous. On the other hand, the partial derivatives of $u$ and $v$ may very well exist even if $f$ is not differentiable. In fact, the partial derivatives may exist even if $f$ is not even continuous.

Recall that differentiabilty for functions of several variables is not the same as existence of (partial) derivatives.

A good example to have is $$ f(z) = \begin{cases} \exp(-1/z^4), & z \neq 0 \\ 0, & z = 0 \end{cases}. $$ Clearly $f$ is holomorphic except at $z=0$ where it has an essential singularity; in particular $f$ is not continuous at $z=0$. Nevertheless, you can check that the partial derivatives of $u$ and $v$ exist everywhere and they satisfy Cauchy-Riemann's equations everywhere, including at $z = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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