Combination Theorem for Continuous Functions/Complex/Sum Rule
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Theorem
Let $\C$ denote the complex numbers.
Let $f$ and $g$ be complex functions which are continuous on an open subset $S \subseteq \C$.
Then:
- $f + g$ is continuous on $S$.
Proof
We have:
\(\ds \forall c \in S: \, \) | \(\ds \lim_{x \mathop \to c} \map f z\) | \(=\) | \(\ds \map f c\) | Definition of Continuous Complex Function | ||||||||||
\(\ds \forall c \in S: \, \) | \(\ds \lim_{x \mathop \to c} \map g z\) | \(=\) | \(\ds \map g c\) | Definition of Continuous Complex Function |
Let $f$ and $g$ tend to the following limits:
\(\ds \lim_{z \mathop \to c} \map f z\) | \(=\) | \(\ds l\) | ||||||||||||
\(\ds \lim_{z \mathop \to c} \map g z\) | \(=\) | \(\ds m\) |
From the Sum Rule for Limits of Complex Functions, we have that:
- $\ds \lim_{z \mathop \to c} \paren {\map f z + \map g z} = l + m$
So, by definition of continuous again, we have that $f + g$ is continuous on $S$.
$\blacksquare$