The complex logarithm

Complex functions and polynomials: Complex functions

The complex logarithm

In our study of the complex exponential function, we found a first example of a function that maps a complex number into another complex number. This time around, we will look at how to define its inverse, the complex logarithm function. Because the complex exponential function is not injective, that is, we can have #\e^{a\cdot\complexi}=\e^{b\cdot\complexi}# without necessarily having #a=b#, we will first set up some general considerations about complex functions.

Before studying complex numbers, we encountered functions such as #f(x)=x^2# that map a real number, #x#, into another real number, #x^2#. One way to stipulate that #f# is a function from #\mathbb{R}# to #\mathbb{R}# is as \[f: \mathbb{R} \rightarrow \mathbb{R}\] where the first instance of #\mathbb{R}# denotes the domain of #f# and the second the codomain, that is, the space to which #x# is mapped. This notation can be applied to functions that are not just from real numbers to real numbers.

Complex functions

A complex function is a function #f# with the complex plane #\mathbb{C}# as its domain and codomain,
\[f: \mathbb{C}\rightarrow \mathbb{C}\]

Example

The exponential function \[f(\blue{z})=\e^{\blue{z}}\]

Real-world applications

One of the most commonly used complex functions is the complex exponential, particularly in the field of Fourier analysis, where complex exponentials are used to decompose signals, such as musical notes, into their constituent frequencies.

Complex functions are also used to describe electric circuits, particularly alternate current (AC) circuits, which include those present in most daily applications. For instance, the propagation constant of a wave describing the voltage of an AC circuit is defined with a complex logarithm.

Codomain, range, and image

The codomain of a function is not the same as the often-mentioned image or range of the function. The codomain refers to the space to which #x# is mapped, while the image or the range of #f# is the set of points of the codomain that are actually reached by the function. In our example of #f(x)=x^2#, the codomain is #\mathbb{R}# while the image is #\left\{ x \in \mathbb{R} | x\geq 0 \right\}#, or in other words, all #x\geq 0#.

There are functions whose domain and codomain are spaces other than #\mathbb{R}# or #\mathbb{C}#, but we will not consider them at this stage.

Let us now define the complex logarithm function.

Complex logarithm

Let #\blue{z}=\green{r}\cdot \e^{\orange{\varphi}\cdot\complexi}# be a complex number in polar-exponential form (the argument #\orange{\varphi}# is the principle value, and therefore #-\pi<\orange{\varphi}\leq \pi#).

We define the complex natural logarithm of #\blue{z}# as
\[ \ln(\blue{z})=\ln(\green{r}) + \orange{\varphi}\cdot\complexi\]

If #\blue{z}# is written with a polar coordinate #\orange{\theta}# satisfying either #\orange{\theta}\leq -\pi# or #\orange{\theta}>\pi#, we need to re-write #\blue{z}# in such a way that #-\pi<\orange{\theta}\leq \pi# before computing its logarithm.

Examples

\[\begin{array}{rcl}
\blue{z}&=&\displaystyle \green{2}\cdot \e^{\orange{\frac{\pi}{2}}\cdot \ii}\\
\ln\left(\blue{z}\right)&=&\displaystyle \ln\left(\green{2}\right)+\orange{\frac{\pi}{2}}\cdot \ii \\ \\
\blue{z}&=& \displaystyle \green{5}\cdot\e^{\orange{\frac{3\pi}{2}}\cdot \ii}\\
&=&\displaystyle \green{5}\cdot \e^{\orange{-\frac{\pi}{2}}\cdot \ii}\\
\ln\left(\blue{z}\right)&=&\displaystyle\ln\left(\green{5}\right)\orange{-\frac{\pi}{2}}\cdot \ii
\end{array}\]

This video shows how to compute the complex natural logarithm of a complex number #\blue{z}=\green{r}\cdot \e^{\orange{\varphi}\cdot\complexi}# given in polar-exponential form as \[\ln(\blue{z})=\ln(\green{r}) + \orange{\varphi}\cdot\complexi\]

It ends with stating computational rules for complex logarithm functions.

The voice in the video is AI-generated and not a human voice.

Domain of the complex logarithm

Now, #\orange{\theta}# is only uniquely defined up to a multiple of #2\pi#. If we had defined the logarithm without restricting ourselves to the principal value of #\blue{z}#, then it would not have been a function, as the following example shows.

Let #\blue{z}= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\cdot \complexi#.
In polar coordinates, we have #\green{r}=\green{1}# and #\orange{\theta}=\orange{\frac{\pi}{4}}+2\cdot k\cdot \pi# for every #k\in\mathbb{Z}#. Thus, we would have \[\ln(\blue{z})=\left(\orange{\frac{\pi}{4}}+2\cdot k\cdot \pi\right)\cdot\complexi \] This corresponds to an infinite set of imaginary numbers for all possible values of #k\in\mathbb{Z}#. If, on the other hand, we use the principal value, then only #k=0# would be acceptable and \[\ln(\blue{z})=\orange{\frac{\pi}{4}}\cdot\complexi\]

Having defined the logarithm, we can look at some of its properties.

Computational rules for complex logarithms

Let #\blue{z}# and #\purple{w}# be complex numbers, while #n# is a natural number. The following three properties apply to the complex logarithm function.

#\ln\left(\blue{z}\cdot \purple{w}\right)=\ln\left(\blue{z}\right)+\ln\left(\purple{w}\right)#
where the principal value #\mathrm{Arg}(\blue{z}\cdot \purple{w})# is in the interval #\ivco{-\pi}{\pi}#, meaning if #\mathrm{Arg}(\blue z)+\mathrm{Arg}(\purple w)# is outside the interval, we must subtract (or add) #2\cdot \pi# to their sum.
#\ln\left(\blue{z}^n\right)=n\cdot\ln\left(\blue{z}\right)#
where, once again, we need to be careful with the arguments of both sides of this equation.
#\e^{\ln\left(\blue{z}\right)}=\ln\left(\e^{\blue{z}}\right)=\blue{z}#

Example 1

Let #\blue{z}=\blue{2 \cdot \e^{\frac{\pi}{2}\cdot \ii}}# and #\purple{w}=\purple{4 \cdot \e^{\frac{\pi}{3}\cdot \ii}}#.
\[\begin{array}{rcl}
\ln(\blue z)+\ln(\purple w) &=& \ln(\blue z \cdot \purple w) \\
&& \quad\blue{\text{rule}}\\
\ln\left(\blue{2\cdot \e^{\frac{\pi}{2}\cdot \ii}}\right) + \ln\left(\purple{4\cdot \e^{\frac{\pi}{3}\cdot \ii}}\right) &=&
\ln\left( \left(\blue{2\cdot \e^{\frac{\pi}{2}\cdot \ii}}\right) \cdot \left(\purple{4\cdot \e^{\frac{\pi}{3}\cdot \ii}}\right) \right) \\
&& \quad\blue{\text{substituted } z \text{ and } w}\\
&=& \ln\left(\green{8}\cdot\e^{\orange{\frac{5\cdot \pi}{6}}\cdot \ii}\right)\\
&&\quad\blue{\text{calculated the product}}\\
&=&\displaystyle\ln\left(\green{8}\right)+\orange{\frac{5\cdot \pi}{6}}\cdot \ii \\
&&\quad\blue{\text{applied the definition of the complex logarithm}}
\end{array}\]

Example 2

Let #\blue{z}=\green{2}\cdot\e^{\orange{-\frac{\pi}{4}}\cdot \ii}#.
\[\begin{array}{rcl}
\ln\left(\blue{z}^8\right)&=&8\cdot\ln\left(\blue{z}\right)\\
&&\quad\blue{\text{used rule}\ln\left(z^n\right)=n\cdot\ln\left(z\right)}\\
\ln\left( \left( \green{2}\cdot\e^{\orange{-\frac{\pi}{4}}\cdot \ii} \right)^8 \right)&=&8 \cdot \ln\left(\green{2}\cdot\e^{\orange{-\frac{\pi}{4}}\cdot \ii}\right)\\
&&\qquad\blue{\text{substituted the definition of }z}\\
&=& 8 \cdot \left( \ln\left(\green{2}\right)\orange{-\frac{\pi}{4}}\cdot\ii \right) \\
&&\quad\blue{\text{applied definition of the complex logarithm}}\\
&=& 8 \cdot \ln\left(\green{2}\right) -2\cdot\pi \cdot\ii \\
&&\quad\blue{\text{expanded brackets}}
\end{array}\]

Example 3

Let #\blue z = \blue{\sqrt{3} \cdot \e^{\frac{\pi}{12}\cdot \ii} }#. Then,
\[\begin{array}{rcl}
\e^{\ln\left(\blue{z}\right)}&=&\blue{z} \\
&&\quad\blue{\text{rule}}\\
\e^{\ln\left( \blue{\sqrt{3} \cdot \e^{\frac{\pi}{12}\cdot \ii} } \right)} &=& \blue{\sqrt{3} \cdot \e^{\frac{\pi}{12}\cdot \ii} }\\
&&\quad \blue{\text{substituted }z}\\
\end{array}\]

Let #\blue z = \blue{ \sqrt{3}+ \ii}#.
\[\begin{array}{rcl}
\ln\left(\e^{\blue{z}}\right)&=&\blue{z}\\
&&\quad \blue{\text{rule}}\\
\ln\left(\e^{\blue{\sqrt{3}+5 \cdot \ii}} \right) &=& \blue{\sqrt{3}+ \ii}\\
&&\quad \blue{\text{substituted }z }\\
\end{array}\]

Proofs

Proof of #\ln\left(\blue{z}\cdot \purple{w}\right)=\ln\left(\blue{z}\right)+\ln\left(\purple{w}\right)#.
Let #\blue{z}=\green{r}_{\blue z}\cdot \e^{\orange{\varphi}_{\blue z}\cdot\complexi}# and #\purple{w}=\green{r}_{\purple w}\cdot \e^{\orange{\varphi}_{\purple w}\cdot\complexi}#. Then, #\blue{z}\cdot \purple{w}# is given by \[\blue{z}\cdot \purple{w}=\left(\green{r}_{\blue z}\cdot \green{r}_{\purple w}\right) \cdot\e^{\left(\orange{\varphi}_{\blue z}+\orange{\varphi}_{\purple w}\right)\cdot \ii}\]Applying the complex logarithm to #\blue{z}\cdot \purple{w}# we obtain
\[\begin{array}{rcl}
\ln\left(\blue{z}\cdot \purple{w}\right)&=&\ln\left(\green{r}_{\blue z}\cdot \green{r}_{\purple w}\right)+\left(\orange{\varphi}_{\blue z}+\orange{\varphi}_{\purple w}\right)\cdot \complexi\\
&&\quad\blue{\text{applied the definition of the complex logarithm}}\\
&=&\ln\left(\green{r}_{\blue z}\right)+\ln\left(\green{r}_{\purple w}\right)+\left(\orange{\varphi}_{\blue z}+\orange{\varphi}_{\purple w}\right)\cdot\ii
\\
&&\quad\blue{\text{used that }\ln\left(a\cdot b\right)=\ln\left(a\right)+\ln\left(b\right)\text{ for real numbers }a\text{ and }b}\\
&=&\left(\ln\left(\green{r}_{\blue z}\right) + \orange{\varphi}_{\blue z}\cdot\ii\right)+\left(\ln\left(\green{r}_{\purple w} \right)+\orange{\varphi}_{\purple w}\cdot\ii\right)\\
&&\quad\blue{\text{grouped the contributions of }z\text{ and }w\text{ by themselves}}\\
&=&\ln\left(\blue{z}\right)+\ln\left(\purple{w}\right)\\
&&\quad\blue{\text{applied the definition of the complex logarithm twice}}
\end{array}\]Thus finishing the proof.
Proof of #\ln\left(\blue{z}^n\right) =n\cdot \ln\left(\blue{z}\right)#.
For this proof, we need to use de Moivre's theorem.
Let #\blue{z}=\green{r}\cdot \e^{\orange{\varphi}\cdot\complexi}#. We first note that \[n\cdot\ln\left(\blue{z}\right)=n\cdot \ln\left(\green{r}\right) + n\cdot\orange{\varphi}\cdot\complexi\]Now, to prove the identity,
\[\begin{array}{rcl} \ln\left(\blue{z}^n\right)&=&\ln\left(\green{r}^n\cdot \e^{n \cdot \orange{\varphi}\cdot\complexi}\right)\\
&&\quad\blue{\text{used de Moivre's theorem}}\\
&=&\ln\left(\green{r}^n\right)+\left(n \cdot \orange{\varphi}\right)\cdot\complexi\\
&&\quad\blue{\text{applied the complex logarithm}}\\
&=&n\cdot\ln\left(\green{r}\right)+n\cdot\orange{\varphi}\cdot\complexi\\
&&\quad\blue{\text{used that }\ln\left(a^n\right)=n \cdot \ln\left(a\right)\text{ for real numbers}}\\
&=&n\cdot\ln\left(\blue{z}\right)\\&&\quad\blue{\text{definition of the complex logarithm }\ln(z)=\ln\left(r\right)+\varphi\cdot \complexi}
\end{array}\]Thus finishing the proof.
Proof #\e^{\ln\left(\blue{z}\right)}=\ln\left(\e^{\blue{z}}\right)=\blue{z}#.
Let #\blue{z}=\green{r}\cdot \e^{\orange{\varphi}\cdot\complexi}#. We first prove that composing the exponential with the logarithm returns #\blue{z}#,
\[\begin{array}{rcl}\e^{\ln\left(\blue{z}\right)}&=&\e^{\ln\left(\green{r}\right)+\orange{\varphi}\cdot \complexi}\\
&&\quad\blue{\text{applied the definition of the logarithm of }z}\\
&=& \e^{\ln\left(\green{r}\right)}\cdot \left( \cos(\orange{\varphi}) + \sin(\orange{\varphi})\cdot \ii \right) \\
&&\quad\blue{\text{applied the definition of the complex exponential}}\\
&&\quad\blue{\text{with }\mathrm{Re}\left(\ln(z)\right) = \ln(r) \text{ and } \mathrm{Im}\left(\ln(z)\right) =\varphi}\\
&=&\e^{\ln\left(\green{r}\right)}\cdot\e^{\orange{\varphi}\cdot\complexi}\\
&&\quad\blue{\text{applied Euler's formula}}\\
&=&\green{r}\cdot \e^{\orange{\varphi}\cdot\complexi}\\
&&\quad\blue{\text{used that }\e^{\ln\left(a\right)}=a\text{ for real numbers}}\\
&=&\blue{z}
\end{array}\] and follow with the converse, i.e., composing the logarithm with the exponential returns #\blue{z}#,
\[\begin{array}{rcl}
\ln\left(\e^{\blue{z}}\right)&=&\ln \left( \e^{\green{r}\cdot (\cos\left(\orange{\varphi}\right) + \sin\left(\orange{\varphi}\right)\cdot \ii } \right) \\
&&\quad \blue{\text{used the polar form of }z}\\
&=&\ln\left(\e^{\green{r}\cdot\cos\left(\orange{\varphi}\right)}\cdot\e^{\green{r}\cdot\sin\left(\orange{\varphi}\right)\cdot\complexi}\right)\\
&&\quad\blue{\text{used the rule } \e^{z+w}=\e^z \cdot \e^w}\\
&=&\ln\left(\e^{\green{r}\cdot\cos\left(\orange{\varphi}\right)}\right)+\green{r}\cdot \sin\left(\orange{\varphi}\right)\cdot\complexi\\
&&\quad\blue{\text{applied the definition of the complex logarithm } }\\
&=&\green{r}\cdot\cos\left(\orange{\varphi}\right)+\green{r}\cdot\sin\left(\orange{\varphi}\right)\cdot\complexi\\
&&\quad\blue{\text{used }\ln\left(\e^a\right)=a\text{ for real }a}\\
&=&\green{r}\cdot \e^{\orange{\varphi}\cdot\complexi}\\
&&\quad\blue{\text{converted to polar-exponential form}}\\
&=&\blue{z}
\end{array}\]Thus finishing the proof.

Consider the complex number #z= 5 \cdot \ee ^{{{5\cdot \pi}\over{6}} \cdot \ii}#.

Express #\ln(z)# in cartesian form.

#\ln(z)=# #\ln(5) + {{5\cdot \pi}\over{6}} \cdot \ii#

We notice that the complex number #z= 5 \cdot \ee ^{{{5\cdot \pi}\over{6}} \cdot \ii}# has norm #r=5# and angle #\theta = {{5\cdot \pi}\over{6}}#.

Since, #-\pi < \theta \leq \pi#, the principal value of #z# is #\varphi = \theta = {{5\cdot \pi}\over{6}}#.

Knowing the norm and principal value of #z#, we can determine its natural logarithm,

\[ \begin{array}{rcl}
\ln(z) & = & \displaystyle \ln(r) + \varphi \cdot \ii \\
&&\qquad \blue{\text{definition of natural logarithm of }z} \\
& = & \displaystyle \ln(5) + {{5\cdot \pi}\over{6}} \cdot \ii \\
&&\qquad \blue{\text{substituted } r = 5 \text{ and } \varphi = {{5\cdot \pi}\over{6}} } \\
\end{array}\]

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