Complex-Differentiable Function is Continuous/Proof 1
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Theorem
Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set.
Let $f$ be complex-differentiable at $a \in D$.
Then $f$ is continuous at $a$.
Proof
Let $\map {N_r} 0$ denote the $r$-neighborhood of $0$ in $\C$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {N_r} 0 \setminus \set 0$:
- $(1): \quad \map f {a + h} = \map f a + h \paren {\map {f'} a + \map \epsilon h}$
where $\epsilon: \map {N_r} 0 \setminus \set 0 \to \C$ is a complex function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
We rewrite the right hand side of $(1)$ to get:
\(\ds \map f {a + h}\) | \(=\) | \(\ds \map f a + h \map {f'} a + h \map \epsilon h\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f z\) | \(=\) | \(\ds \map f a + \paren {z - a} \map {f'} a + \paren {z - a} \map \epsilon {z - a}\) | substituting $z = a + h$ for $z \in \map {N_r} a$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{z \mathop \to a} \map f z\) | \(=\) | \(\ds \lim_{z \mathop \to a} \paren {\map f a + \paren {z - a} \map {f'} a + \paren {z - a} \map \epsilon {z - a} }\) | Take the limit on both sides | ||||||||||
\(\ds \) | \(=\) | \(\ds \map f a + \lim_{z \mathop \to a} \paren {z - a} \map {f'} a + \lim_{z \mathop \to a} \paren {\paren {z - a} \map \epsilon {z - a} }\) | Combination Theorem for Limits of Complex Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f a\) | as $\ds \lim_{z \mathop \to a} \paren {z - a} = 0$, and $\ds \lim_{z \mathop \to a} \map \epsilon {z - a} = 0$ |
By definition of continuous complex function, it follows that $f$ is continuous at $a$.
$\blacksquare$