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2.4 Continuous Functions

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Continuity of functions of a complex variable is much like continuity of functions of a real variable. In this section we define what it means for a function $f : D \to \C$ on a domain to be continuous.

Definition (Continuity of a Function)

Fix a domain $D \subset \C$. A function $f: D\to\C$ is continuous at a point $c \in D$ if \[\lim_{z \to c} f(z) = f(c)\] holds.

The limit above is formally defined as follows.

Definition (Limits)

Fix a domain $D \subset \C$ and $f: D\to\C$. Fix also $c \in D$. We say that \[\lim_{z\to c} f(z) = \ell\] or that $f(z)$ tends to $\ell$ as $z$ tends to $c$, or that the limit of $f(z)$ exists as $z \to c$ if, for all $\epsilon > 0$, there exists $\delta > 0$ such that if $z \in D$ and $0 < |z-c| < \delta$ then $|f(z)-\ell| < \epsilon$.

That is, $f(z) \to \ell$ as $z \to c$ means that if $z$ is very close and not equal to $c$ then $f(z)$ is very close to $\ell$. Note that in this definition we do not need to know the value of $f(c)$.

Example

Let $f : \C \to \C$ be defined by \[f(z) = \begin{cases} 1 & z \ne 0 \\ 0 & z = 0 \end{cases}\] so that \[\lim\limits_{z\to 0} f(z) = 1\] but $\displaystyle\lim\limits_{z\to 0}f(z) \ne f(0)$. Thus $f$ is not continuous at 0.

Continuous functions of a complex variable obey the same rules as continuous functions of a real variable. Therefore we can use the same strategies for testing and evaluating limits in complex analysis as we used in real analysis.

Lemma (Limit Laws)

Suppose that $\displaystyle\lim_{z \to c} f(z) = \ell$ and $\displaystyle\lim_{z \to c} g(z) = m$.

  1. $\displaystyle\lim_{z \to c} f(z) + g(z) = \ell + m$
  2. $\displaystyle\lim_{z \to c} a f(z) = a \ell$ for any $a \in \C$
  3. $\displaystyle\lim_{z \to c} f(z) g(z) = \ell m$
  4. $\displaystyle\lim_{z \to c} f(z) / g(z) = \ell / m$ if $m \ne 0$
Lemma (Polynomial Limits)

If \[f(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0\] then $f$ is continuous at every $c \in \C$.

Lemma (Rational Limits)

If $f,g$ are polynomials and $g(c) \ne 0$ then \[\lim_{z \to c} \dfrac{f(z)}{g(z)} = \dfrac{f(c)}{g(c)}\]