Complex numbers: Complex functions
The complex logarithm
Finally, we introduce the complex logarithm. Here we want, as discussed in the beginning, leave as much as possible properties of real functions intact. In particular, we want to keep the calculation rule \[\ln(z)=\ln({r\cdot \e^{\varphi\cdot\ii}})=\ln(r)+ \varphi\cdot\ii\] for \(z=r\cdot \e^{\varphi\cdot\ii}\) for a complex number equal to \(0\) in polar form.
This suggests a definition for the complex logarithm \(\ln(z)\), but there is a snag: to \(\varphi\) we can add any integral multiple of \(2\pi\) without changing the value of \(z\). So we have \(\ln(z)=\ln(r)+ \mathrm{i}(\varphi+2k\pi)\) for integers \(k\). The logarithm is thus an example of a multi-valued function, i.e., a function that can take multiple values for each function argument. We can of course choose a specific value: say \(k=0\) under the assumption that \(-\pi<\varphi\le \pi\). In other words, we take the principal value of the argument \(z\). That's why we call this the principal value of \(\boldsymbol{\ln(z)}\) and speak of the principal branch of the complex logarithm when we do this for all originals under the mapping \(z\mapsto\ln(z)\).
\[\begin{aligned} \ln(z) &= \ln(|z|) + \mathrm{i}\,\mathrm{arg}(z)\\ \\ \text{The principal branch of }\ln(z) &=\ln(|z|) + \mathrm{i}\,\mathrm{Arg}(z)\end{aligned}\]
In some math books one uses the notation \(\mathrm{Ln}\) for the principal branch of the complex logarithm, but we will not use this notation.
The inverse of the complex exponential function
- The complex exponential function \(\exp\) maps every original \(z\) onto a unique image if \(-\pi\lt \mathrm{Im}(z)\le \pi\). More formally, we call the complex exponential function \(\exp\) injective on the domain \(\left\{z\in\mathbb{C}\mid -\pi\lt \mathrm{Im}(z)\le\pi\right\}\) .
- The range of \(\exp\) in this domain is the set of all complex numbers unequal to \(0\). Thus, the range is equal to \(\mathbb{C}\setminus\{0\}\).
- The principal branch of the complex logarithm \(\ln\) is the inverse function of the complex exponential function limited to \(\left\{z\in\mathbb{C}\mid -\pi\lt \mathrm{Im}(z)\le\pi\right\}\). So, the domain of the principal branch of \(\ln\) is the set of all complex numbers unequal to #0# and the range consists of all complex numbers \(z\) with \(-\pi\lt \mathrm{Im}(z)\le \pi\).
For the maths enthusiast we give the proof.
Let \(z\) and \(w\) be complex numbers with \(\exp(z)=\exp(w)\). Then \(e^z=\e^w\) and because of the calculation rules for complex powers \(\e^{z-w}=1\). The definition of the complex exponential function teaches us that \[\begin{aligned}e^{\mathrm{Re}(z-w)} &=\left|\e^{z-w}\right|=\left| 1\right|=1\\ \mathrm{Im}(z-w) &=\arg\left(e^{z-w}\right)=0\pmod{2\,\pi}\end{aligned}\] Thus \[z-w=\mathrm{Re}(z-w)+\mathrm{Im}(z-w)\cdot\ii=0\pmod{2\,\pi\,\ii}\] Thus \[z=w\pmod{2\,\pi\,\ii}\] If \(z\) and \(w\) are both in the domain \(\left\{z\in\mathbb{C}\mid -\pi\lt \mathrm{Im}(z)\le\pi\right\}\), then their imaginary parts differ less than \(2\,\pi\) from each other, so that they must be equal. This proves \(z=w\), and thus the injectivity of \(\exp\) as a function on the given domain.
It is known that each complex number different from \(0\) has a polar form and can therefore be written as \(e^z\) for a suitable \(z\) with \(-\pi\lt \mathrm{Im}(z)\le \pi\). This means that the image of \(\exp\) on the given domain is \(\mathbb{C}\setminus\{0\}\).
The fact that \(\ln\) is the inverse of \(\exp\) on the given domain, finally follows from: \[\begin{aligned}\ln\left(e^z\right) &=\ln\left(\left|e^{z}\right|\right) + \arg\left(e^z\right)\cdot\ii\\ &\phantom{uvwxyz}\color{blue}{\text{definition of the complex logarithm}}\\ &=\ln\left(e^{\mathrm{Re}(z)}\right) + \mathrm{Im}(z)\cdot\ii\\ &\phantom{uvwxyz}\color{blue}{\text{definition of the complex function }\exp}\\ &=\mathrm{Re}(z) + \mathrm{Im}(z)\,\ii\\ &=z\end{aligned}\]
We mention some calculation rules that are similar to rules of the real natural logarithm, but often with small differences:
Calculation rules for the complex logarithm Let \(z\) and \(w\) be two nonzero complex numbers and let \(n\) be a natural number. Then:
- #\ln\left(z\cdot w\right)=\ln\left(z\right)+\ln\left(w\right)\pmod{2\,\pi\,\ii}#
- #\ln(z^n)=n\cdot \ln\left(z\right)\pmod{2\,\pi\,\ii}#
- #e^{\ln\left(z\right)}=z#
- #\ln\left(e^{z}\right)=z\pmod{2\,\pi\,\ii}#
For the maths enthousiast, we give proofs of the four equalities.
The first equality follows from \[\begin{aligned}\ln\left(z\cdot w\right) &= \ln\left(\left|z\cdot w\right|\right)+\arg\left(z\cdot w\right)\cdot\ii\\
&= \ln\left(\left|z\right|\right)+\ln\left(\left| w\right|\right)+\arg\left(z\right)\cdot\ii+\arg\left(w\right)\cdot\ii\pmod{2\,\pi\,\ii}\\
&=\ln(z)+\ln(w)\pmod{2\,\pi\,\ii}\end{aligned}\] The second equality follows from repeated application of the first with \(z^k=z^{k-1}\cdot z\) for \(k=n,n-1,\ldots,2\) .
The third equality follows from the fact that the logarithm is the inverse of \(\exp\).
For \(z\) with \(-\pi\lt \mathrm{Im}(z)\le\pi\) follows the fourth equality also from the fact that the logarithm is the inverse of \(\exp\). For other values of \(z\) you can not draw this conclusion because \(z\) then does not come from the specially chosen domain of \(\exp\) in which the function has a unique image for each original. However, direct calculation shows that the complex logarithm modulo \(2\pi\cdot\ii\) behaves as the inverse function: \[\begin{aligned}\ln\left(e^{z}\right)&= \ln\left(\left|e^{z}\right|\right)+\arg\left(e^{z}\right)\cdot\ii\\
&\phantom{uvwxyz}\color{blue}{\text{definition of the complex logarithm}}\\
&= \ln\left(e^{\mathrm{Re}(z)}\right)+\bigl(\mathrm{Im}(z) \pmod{2\,\pi}\bigr)\cdot\ii\\
&\phantom{uvwxyz}\color{blue}{\text{definition of the complex function }\exp}\\
&= \left(\mathrm{Re}(z)+\mathrm{Im}(z)\cdot\ii\right) \pmod{2\,\pi\,\ii}\\
&\phantom{uvwxyz}\color{blue}{\text{definition of the real logarithm}}\\
&= z\pmod{2\,\pi\,\ii}\end{aligned}\]
In the above calculation rules, the addition of modulo \((\mathrm{mod\;}2\,\pi\,\ii)\) is necessary. Bear this in mind or otherwise you will quickly obtain conflicting results
Through the complex exponential function, and possibly a calculator, you can calculate function values. We give a few examples.
If \[z=r\, e^{\ii\,\phi}\] then \[z=\ln(r)+\ii\,\phi\quad(\text{mod }2\,\pi\,\ii)\] In this sum: \[1+\sqrt{3}\,\ii=2\,e^{\tfrac{1}{3}\pi\,\ii}\] So \[z=\ln(2)+\tfrac{1}{3}\pi\,\ii\quad (\mathrm{mod\;} 2\,\pi\,\ii)\]
