For complex function : $f'(z)$ exists $\implies$ $f$ continous in $z$?

Question

Consider a complex function $f(z): A\subset\mathbb C \to\mathbb C$.

If the derivative of $f$ exists then $f$ must necessarily be a continuous function?

Is the following true? $f'(z)$ exists $\implies$ $f$ continuous in $z$

Answer 1

A quick proof: Let $f:A\to\mathbb C$ and let $z_0\in A$ be an accumulation point of $A$. Suppose $f'(z_0)$ exists, i.e. the limit

$$\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=L$$

exists and $L\in\mathbb C$. Then we have

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

Thus, $\lim_{z\to z_0}f(z)=f(z_0)$ and $f$ is continuous in $z_0$.