# Is the condition of continuity for the differentable functions necessary in Looman-Menchoff theorem?

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?

• You might appreciate this Monthly article. – user98602 Dec 27 '14 at 21:16

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.
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 $z = 0$.