A (complex) function\(f\) is a map from a subset \(G \subseteq \C\) to \(\C\text{;}\) in this situation we will write \(f : G \to \C\) and call \(G\) the domain of \(f\text{.}\) This means that each element \(z \in G\) gets mapped to exactly one complex number, called the image of \(z\) and usually denoted by \(f(z)\text{.}\)
So far there is nothing that makes complex functions any more special than, say, functions from \(\R^m\) to \(\R^n\text{.}\) In fact, we can construct many familiar looking functions from the standard calculus repertoire, such as \(f(z)=z\) (the identity map), \(f(z)=2z+i\text{,}\)\(f(z)=z^3\text{,}\) or \(f(z) = \frac 1 z\text{.}\) The former three could be defined on all of \(\C\text{,}\) whereas for the latter we have to exclude the origin \(z=0\) from the domain. On the other hand, we could construct some functions that make use of a certain representation of \(z\text{,}\) for example, \(f(x,y) = x-2iy\text{,}\)\(f(x,y) = y^2-ix\text{,}\) or \(f(r,\phi) = 2r \, e^{ i (\phi + \pi)
}\text{.}\)
Next we define limits of a function. The philosophy of the following definition is not restricted to complex functions, but for sake of simplicity we state it only for those functions.
Suppose \(f: G \to \C\) and \(z_0\) is an accumulation point of \(G\text{.}\) If \(w_0\) is a complex number such that for every \(\epsilon>0\) we can find \(\gd>0\) so that, for all \(z\in G\) satisfying \(0 \lt \left| z - z_0 \right| \lt \gd\text{,}\) we have \(\left| f(z) - w_0 \right| \lt \epsilon\text{,}\) then \(w_0\) is the limit of \(f\) as \(z\) approaches \(z_0\text{;}\) in short,
This definition is the same as is found in most calculus texts. The reason we require that \(z_0\) is an accumulation point of the domain is just that we need to be sure that there are points \(z\) of the domain that are arbitrarily close to \(z_0\text{.}\) Just as in the real case, our definition (i.e., the part that says \(0\lt \abs{z-z_0}\)) does not require that \(z_0\) is in the domain of \(f\) and, if \(z_0\) is in the domain of \(f\text{,}\) the definition explicitly ignores the value of \(f(z_0)\text{.}\)
Given \(\epsilon > 0\text{,}\) we need to determine \(\gd > 0\) such that \(0 \lt |z-i| \lt \gd\) implies \(|z^2 + 1| \lt
\epsilon\text{.}\) We rewrite
If we choose \(\gd\text{,}\) say, smaller than 1 then the factor \(|z+i|\) on the right can be bounded by 3 (draw a picture!). This means that any \(\gd \lt \min \{ \frac \epsilon 3, 1
\}\) should do the trick: in this case, \(0 \lt |z-i| \lt \gd\) implies
Just as in the real case, the limit \(w_0\) is unique if it exists (see Exercise 2.5.3). It is often useful to investigate limits by restricting the way the point \(z\) approaches \(z_0\text{.}\) The following result is a direct consequence of the definition.
Suppose \(f: G \to \C\) and \(\lim_{z\to z_0}f(z) =
w_0\text{.}\) Suppose \(\widetilde G \subseteq G\) and \(z_0\) is an accumulation point of \(\widetilde G\text{.}\) If \(\widetilde f\) is the restriction of \(f\) to \(\widetilde G\text{,}\) then \(\lim_{z\to z_0} \widetilde
f(z)\) exists and has the value \(w_0\text{.}\)
The definition of limit in the complex domain has to be treated with a little more care than its real companion; this is illustrated by the following example.
To see this, we try to compute this limit as \(z \to 0\) on the real and on the imaginary axis. In the first case, we can write \(z = x \in \R\text{,}\) and hence
\begin{equation*}
\lim_{ z \to 0 } \frac{ \, \overline z \, }{ z } \ = \ \lim_{ x
\to 0 } \frac{ \, \overline x \, }{ x } \ = \ \lim_{ x \to 0 }
\frac{ x }{ x } \ = \ 1 \, \text{.}
\end{equation*}
So we get a different “limit” depending on the direction from which we approach 0. Proposition 2.1.4 then implies that the limit of \(\frac{\, \overline z \, } z\) as \(z \to 0\) does not exist.
On the other hand, the following usual limit rules are valid for complex functions; the proofs of these rules are everything but trivial and make for nice practice (see Exercise 2.5.4); as usual, we give a sample proof.
Let \(f\) and \(g\) be complex functions with domain \(G\text{,}\) let \(z_0\) be an accumulation point of \(G\text{,}\) and let \(c \in \C\text{.}\) If \(\lim_{ z \to z_0 } f(z)\) and \(\lim_{ z \to z_0 } g(z)\) exist, then
\(\displaystyle
\lim_{ z \to z_0 } \left( f(z) + c g(z) \right)
= \lim_{ z \to z_0 } f(z) + c \lim_{ z \to z_0 } g(z) \,
\text{.}\)
We will prove (a). Assume that \(c \ne 0\) (otherwise there is nothing to prove), and let \(L = \lim_{ z \to z_0 } f(z)\) and \(M = \lim_{ z
\to z_0 } g(z)\text{.}\) Then we know that, given \(\epsilon > 0\text{,}\) we can find \(\delta_1, \delta_2 > 0\) such that
Here we used the triangle inequality (Proposition 1.3.4). This proves that \(\lim_{ z \to z_0 } (f(z) + c \, g(z)) = L+c
\, M\text{,}\) which was our claim.
then \(f\) is continuous at \(z_0\text{.}\) More generally, \(f\) is continuous on \(E \subseteq G\) if \(f\) is continuous at every \(z \in E\text{.}\)
However, in almost all proofs using continuity it is necessary to interpret this in terms of \(\epsilon\)’s and \(\delta\)’s. So here is an alternate definition:
Suppose \(f: G \to \C\) and \(z_0 \in G\text{.}\) Then \(f\) is continuous at \(z_0\) if, for every positive real number \(\epsilon\text{,}\) there is a positive real number \(\delta\) so that
\begin{equation*}
|\,f(z)-f(z_0)\,| \lt \epsilon \;\text{ for all } \; z\in G
\; \text{ satisfying } \; |\,z-z_0\,| \lt \delta\, \text{.}
\end{equation*}
We already proved (in Example 2.1.3) that the function \(f: \C \to \C\) given by \(f(z) = z^2\) is continuous at \(z = i\text{.}\) You’re invited (see Exercise 2.5.8) to extend our proof to show that, in fact, this function is continuous on \(\C\text{.}\)
The image of the function \(g: G \to \C\) is the set \(\left\{ g(z) : \, z \in G \right\}
\text{.}\) If the image of \(g\) is contained in the domain of another function \(f : H \to \C\text{,}\) we define the composition\(f \circ g : G \to \C\) through
Let \(g : G \to \C\) with image contained in \(H\text{,}\) and let \(f : H \to \C\text{.}\) Suppose \(z_0\) is an accumulation point of \(G\text{,}\)\(\lim_{z\to z_0}g(z)=w_0\in H\text{,}\) and \(f\) is continuous at \(w_0\text{.}\) Then \(\lim_{z\to z_0}f(g(z))=f(w_0)\text{;}\) in short,
