Complex Analysis – Is There a Function Differentiable at z_0 but Not Analytic?

Question

Recently, I am learning complex analysis using " Complex Analysis for mathematics and engineering" by John H. Mathews and Russell W. Howell.

The definition of analytic function at a point $z_0$ is defined as follows:

If a function $f(z)$ is differentiable at $z_0$ and its neighborhood, then we say that $f$ is analytic at $z_0$.

But nearly every nice function e.g. $\sin z$, $\text{exp}(z)$ is analytic and the professor doesn't even check that in the neighborhood of $z_0$, $f$ is analytic.

So, my question is:

Is analytic function the same as a differentiable function in complex analysis? Or is there any counterexample?

Thank you!

Answer 1

The first big result in complex analysis is that "a function is holomorphic if and only if it is analytic":

A function $f:U \to \mathbb{C}$ is holomorphic if $U$ is open and $f$ is differentiable.

A function $f: A \to \mathbb{C}$ is analytic at $p \in A$ if it can be written as a power series in a neighbourhood of $p$. The function is analytic if it is analytic at every point in its domain.

Thus if $f$ is analytic at every point in its domain, the domain is automatically open. However, "differentiability" doesn't require openness.

The nice functions you have outlined are known to be differentiable everywhere. This means if you restrict the domain to any open subset of $\mathbb C$, it'll still be differentiable on an open set, so it is holomorphic (and analytic). However, if your function is "not nice", it may be differentiable on a closed set that cannot be extended to an open set.

$f(z) = |z|^2$ is a famous example. If you calculate the Cauchy-Riemann equations, you will see that it is only differentiable at the point $z = 0$. This is not differentiable on any nonempty open set, so it is not analytic.