# Showing a certain function is analytic given only continuity.

Let $\Omega$ be an open subset of $\mathbb{C}$ and $z_0\in\Omega$. Show that a function $f:\Omega\rightarrow \mathbb{C}$ is analytic at $z_0$ iff

$$g(z) = \begin{cases} \dfrac{f(z)-f(z_0)}{z-z_0}-f'(z_0) & \text{if z\neq z_0}\\[6px] 0 & \text{if z=z_0} \end{cases}$$ is continuous at $z_0$.

I think the forward direction is clear enough, but I am at a loss for a strategy for the reverse direction. I am new to the subject and would like to approach the problem from the ground level" if possible.

• $g$ is continuous at $z_0$ if and only if $f$ is complex differentiable at $z_0$, it need not be analytic at $z_0$. ($f$ is analytic at $z_0$ if it is representable by a power series on some neighbourhood of $z_0$.) – Daniel Fischer Feb 18 '16 at 17:15
• What do you mean by "analytic at $z_0\>$"? Maybe you have an $f$ in mind which is analytic in $\Omega\setminus\{z_0\}$ to start with. – Christian Blatter Feb 18 '16 at 17:16
• Isn't $|z|^2$ a counter-example? It is complex differentiable at $0$, thus $g$ is continuous at 0, but it is not analytic. – johnnycrab Feb 18 '16 at 17:23
• Christian Blatter- Yes, I think "analytic at $z_0$" means as you say. – Earl Feb 18 '16 at 18:22
• In my course we have not yet been introduced to the power series representation perspective yet. It may be that we are abusing the the definition of analytic in this exercise. – Earl Feb 18 '16 at 19:04

## 1 Answer

As shown above, $f$ need not be analytic at $z_0$, but the statement is true assuming $f$ is complex differenciable.

Assume first that $f$ is analytic at $z_0$. Then

$$\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$$ exists, and the limit equals to $f'(z_0)$, so

$$\left( \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\right)-f'(z_0)=\lim_{z\to z_0}\left(\frac{f(z)-f(z_0}{z-z_0}-f'(z_0)\right)=0$$

Hence, $g$ is continous at $z_0$.

The other implication follows from the fact that since $g$ is continous at $z_0$, then

$$\lim_{z\to z_0}\left(\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)\right)=0$$ So $$\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=f'(z_0)$$

Therefore, $f$ is complex differentiable at $z_0.$