Definition 3.5.1.
Given a region \(G\text{,}\) any continuous function \(\fLog : G \to \C\) that satisfies \(\exp(\fLog z) = z\) is a branch of the logarithm (on \(G\)).
Section 3.5 Logarithms and Complex Exponentials
The complex logarithm is the first function we’ll encounter that is of a somewhat tricky nature. It is motivated as an inverse to the exponential function, that is, we’re looking for a function \(\fLog\) such that
\begin{equation}
\exp ( \fLog(z) ) \ = \ z \ = \ \fLog ( \exp z ) \,\text{.}\tag{3.3}
\end{equation}
But because \(\exp\) is not one-to-one, this is too much to hope for. In fact, given a function \(\fLog\) that satisfies the first equation in (3.3), the function \(f(z) = \fLog(z) + 2 \pi i\) does as well, and so there cannot be an inverse of \(\exp\) (which would have to be unique). On the other hand, \(\exp\) becomes one-to-one if we restrict its domain, so there is hope for a logarithm if we are careful about its construction and about its domain.
To make sure this definition is not vacuous, let’s write, as usual, \(z = r \, e^{i \phi}\text{,}\) and suppose that \(\fLog z = u(z) + i \, v(z)\text{.}\) Then for the first equation in (3.3) to hold, we need
\begin{equation*}
\exp ( \fLog z ) \ = \ e^{u} e^{i v} \ = \ r \, e^{i \phi} \ = \ z \,\text{,}
\end{equation*}
that is, \(e^{u} = r\) and \(e^{i v} = e^{i \phi}\text{.}\) The latter equation is equivalent to \(v = \phi + 2 \pi k\) for some \(k \in \Z\text{,}\) and denoting the natural logarithm of the positive real number \(x\) by \(\ln(x)\text{,}\) the former equation is equivalent to \(u = \ln |z|\text{.}\) A reasonable definition of a logarithm function \(\fLog\) would hence be \(\fLog z = \ln |z| + i \fArg z\) where \(\fArg z\) gives the argument for the complex number \(z\) according to some convention—here is an example.
Definition 3.5.2.
Let \(\Arg z\) denote the unique argument of \(z \ne 0\) that lies in \((-\pi, \pi]\) (the principal argument of \(z\)). Then the principal logarithm is the function \(\Log: \C \setminus \{ 0 \} \to \C\) defined through
\begin{equation*}
\Log(z) := \ln |z| + i \Arg(z) \,\text{.}
\end{equation*}
Example 3.5.3.
Here are a few evaluations of \(\Log\) to illustrate this function:
\begin{align*}
\Log(2) \amp \ = \ \ln(2) + i \Arg(2) \ = \ \ln(2)\\
\Log(i) \amp \ = \ \ln(1) + i \Arg(i) \ = \ \frac{ \pi i }{
2 }\\
\Log(-3) \amp \ = \ \ln(3) + i \Arg(-3) \ = \ \ln(3) + \pi
i\\
\Log(1-i) \amp \ = \ \ln( \sqrt 2 ) + i \Arg(1-i) \ = \
\frac 1 2 \ln(2) - \frac{ \pi i }{ 4 } \, \text{.}
\end{align*}
SageMath computes “our” principal logarithm.
The principal logarithm is not continuous on the negative part of the real line, and so \(\Log\) is a branch of the logarithm on \(\C \setminus
\R_{ \le 0 }\text{.}\) Any branch of the logarithm on a region \(G\) can be similarly extended to a function defined on \(\overline G \setminus \{ 0
\}\text{.}\) Furthermore, the evaluation of any branch of the logarithm at a specific \(z_0\) can differ from \(\Log(z_0)\) only by a multiple of \(2 \pi i\text{;}\) the reason for this is once more the periodicity of the exponential function.
So what about the second equation in (3.3), namely, \(\fLog (\exp z) = z\text{?}\) Let’s try the principal logarithm: if \(z=x+iy\) then
\begin{align*}
\Log ( \exp z ) \amp \ = \ \ln | e^x e^{iy} | + i \Arg ( e^x
e^{iy} )\\
\amp \ = \ \ln e^x + i \Arg ( e^{iy} )\\
\amp \ = \ x + i \Arg ( e^{iy} )\text{.}
\end{align*}
The right-hand side is equal to \(z=x+iy\) if and only if \(y
\in (-\pi, \pi]\text{.}\) Something similar will happen with any other branch \(\fLog\) of the logarithm—there will always be many \(z\)’s for which \(\fLog ( \exp z ) \ne z\text{.}\)
A warning sign pointing in a similar direction concerns the much-cherished real logarithm rule \(\ln(xy) = \ln(x) +
\ln(y)\text{;}\) it has no analogue in \(\C\text{.}\) For example,
\begin{equation*}
\Log(i) + \Log(i-1) \ = \ i \, \tfrac \pi 2 + \ln \sqrt 2 + \tfrac
{3 \pi i} 4 \ = \ \tfrac 1 2 \ln 2 + \tfrac {5 \pi i} 4
\end{equation*}
but
\begin{equation*}
\Log(i(i-1)) \ = \ \Log(-1-i) \ = \ \tfrac 1 2 \ln 2 - \tfrac {3 \pi
i} 4 \, \text{.}
\end{equation*}
The problem is that we need to come up with an argument convention to define a logarithm and then stick to this convention. There is quite a bit of subtlety here; e.g., the multi-valued map
\begin{equation*}
\arg z \ := \ \text{ all possible arguments of } z
\end{equation*}
gives rise to a multi-valued logarithm via
\begin{equation*}
\log z \ := \ \ln |z| + i \arg z \,\text{.}
\end{equation*}
Neither \(\arg\) nor \(\log\) is a function, yet \(\exp ( \log z ) = z\text{.}\) We invite you to check this thoroughly; in particular, you should note how the periodicity of the exponential function takes care of the multi-valuedness of \(\log\text{.}\)
To end our discussion of complex logarithms on a happy note, we prove that any branch of the logarithm has the same derivative; one just has to be cautious with regions of holomorphicity.
Proposition 3.5.4.
If \(\fLog\) is a branch of the logarithm on \(G\text{,}\) then \(\fLog\) is differentiable on \(G\) with
\begin{equation*}
\frac{ d }{ \diff{z} } \fLog(z) \ = \ \frac 1 z \,\text{.}
\end{equation*}
Proof.
The idea is to apply Proposition 2.2.8 to \(\exp\) and \(\fLog\text{,}\) but we need to be careful about the domains of these functions. Let \(H := \{ \fLog(z) : \, z \in G \}\text{,}\) the image of \(\fLog\text{.}\) We apply Proposition 2.2.8 with \(f: H \to G\) given by \(f(z) = \exp(z)\) and \(g: G \to H\) given by \(g(z) = \fLog(z)\text{;}\) note that \(g\) is continuous, and Exercise 3.6.47 checks that \(f\) and \(g\) are inverses of each other. Thus Proposition 2.2.8 gives
\begin{equation*}
\fLog'(z) \ = \ \frac 1 { \exp' ( \fLog z ) } \ = \ \frac 1 { \exp (
\fLog z ) } \ = \ \frac 1 z \,\text{.}
\end{equation*}
We finish this section by defining complex exponentials.
Definition 3.5.5.
Given \(a, b \in \C\) with \(a \ne 0\text{,}\) the principal value of \(a^b\) is defined as
\begin{equation*}
a^b \ := \ \exp ( b \Log(a)) \,\text{.}
\end{equation*}
Naturally, we can just as well define \(a^b\) through a different branch of the logarithm; our convention is that we use the principal value unless otherwise stated. Exercise 3.6.50 explores what happens when we use the multi-valued \(\log\) in the definition of \(a^b\text{.}\)
One last remark: it now makes sense to talk about the function \(f(z) = e^z\text{,}\) where \(e\) is Euler’s number and can be defined, for example, as \(e = \lim_{n \to \infty} \left( 1 + \frac 1 n \right)^n\text{.}\) In calculus we can prove the equivalence of the real exponential function (as given, for example, through a power series) and the function \(f(x) = e^x\text{.}\) With our definition of \(a^z\text{,}\) we can now make a similar remark about the complex exponential function. Because \(e\) is a positive real number and hence \(\Arg e = 0\text{,}\)
1
Named after Leonard Euler (1707–1783). Continuing our footnote on p. Footnote 1.2.2 we have now honestly established Euler’s formula \(e^{ 2 \pi i } = 1\text{.}\)
\begin{align*}
e^z \amp \ = \ \exp ( z \Log(e))\\
\amp \ = \ \exp \left( z \left( \ln |e| + i
\Arg(e) \right) \right)\\
\amp \ = \ \exp \left( z \ln(e) \right)\\
\amp \ = \ \exp
\left( z \right)\text{.}
\end{align*}
A word of caution: this only works out this nicely because we have carefully defined \(a^b\) for complex numbers. Using a different branch of logarithm in the definition of \(a^b\) can easily lead to \(e^z \ne \exp(z)\text{.}\)
