Complex-Differentiable Function is Continuous

Theorem
Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set.

Suppose that $f$ is complex-differentiable at $z \in D$.

Then $f$ is continuous at $z$.

Proof
By the Alternative Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:


 * $\displaystyle \lim_{h \to 0} f\left({z + h}\right) = f \left({z}\right) + h \left({f \left({z}\right) + \epsilon \left({h}\right) }\right)$

It follows from definition of limit that for all $\epsilon \in \R_{>0}$, there exists $\delta \in \R_{>0}$ with this property:

If $\left\vert{h - 0}\right\vert < \delta$, then $\left\vert{f \left({z+h}\right) - f \left({z}\right)}\right\vert < \epsilon$.

Put $z' = z+h$, so $\left\vert{z' - z}\right\vert = \left\vert{h}\right\vert$.

It follows that:

If $\left\vert{z' - z}\right\vert < \delta$, then $\left\vert{f \left({z'}\right) - f \left({z}\right)}\right\vert < \epsilon$.

Then $f$ is continuous at $z$ by definition.