Combination Theorem for Continuous Functions/Complex/Quotient Rule

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:

$\dfrac f g$ is continuous on $S \setminus \set {z \in S: \map g z = 0}$

that is, on all the points $z$ of $S$ where $\map g z \ne 0$.

Proof

By definition of continuous:

$\forall c \in S: \ds \lim_{z \mathop \to c} \map f z = \map f c$

$\forall c \in S: \ds \lim_{z \mathop \to c} \map g z = \map g c$

Let $f$ and $g$ tend to the following limits:

$\ds \lim_{z \mathop \to c} \map f z = l$

$\ds \lim_{z \mathop \to c} \map g z = m$

From the Quotient Rule for Limits of Complex Functions, we have that:

$\ds \lim_{z \mathop \to c} \frac {\map f z} {\map g z} = \frac l m$

wherever $m \ne 0$.

So, by definition of continuous again, we have that $\dfrac f g$ is continuous on all points $z$ of $S$ where $\map g z \ne 0$.

$\blacksquare$