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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.7 Logarithmic and trigonometric functions

Subsection Logarithmic functions

When showing that \(\exp\colon \C\rightarrow \C\) has image \(\C-\{0\}\text{,}\) we showed that for any \(z\in \C-\{0\}\text{,}\) the set of solutions to \(e^w=z\) is the set
\begin{equation*} \{w\in \C\colon \Re w=\ln \abs{z} \text{ and } \Im w\in \arg(z)\}=\{\ln\abs{z}+i(\Arg z+2\pi k)\colon k\in \Z\}\text{.} \end{equation*}
This motivates the following definition.

Definition 1.7.1. Logarithm.

Given \(z\in \C-\{0\}\text{,}\) we define the logarithm of \(z\text{,}\) denoted \(\log z\text{,}\) to be the infinite set
\begin{equation*} \log z=\{w\in \C\colon \Re w=\ln \abs{z} \text{ and } \Im w\in \arg(z)\}=\{\ln\abs{z}+i(\Arg z+2\pi k)\colon k\in \Z\}\text{.} \end{equation*}
The principal branch of the logarithm is the function \(\Log\colon \C-(-\infty,0]\rightarrow \C\) defined as
\begin{equation*} \Log z=\ln\abs{z}+i\Arg z \end{equation*}

Remark 1.7.2.

Using the same sort of shorthand as in Remark 1.6.9, we can write
\begin{align*} \log z \amp = \ln\abs{z}+i\arg z\text{,} \end{align*}
where as always
\begin{equation*} \arg z=\{\Arg z+2\pi k\colon k\in \Z\} \end{equation*}
is the set of all arguments of \(z\text{.}\)

Remark 1.7.3. \(\Log\).

We will see in due course that there is a good reason for the funny domain specified for \(\Log z\text{.}\) Namely, although we can in principle define \(\Log z\) for any \(z\ne 0\text{,}\) the domain \(\C-(-\infty, 0]\) ensures that \(\Log z\) is continuous, and is in a sense maximal in this regard.

Proof.

Recall that we remarked in the discussion above Figure 1.6.15 that the given \(f\) is a bijection, and hence invertible; the theorem claims that \(\Log\) is the inverse of this function. First observe that \(\im \Log\subseteq R\text{,}\) since
\begin{align*} w\in \C-(-\infty,0] \amp \implies -\pi< \Arg w < \pi\\ \amp\implies -\pi < \Im(\Log z) < \pi\\ \amp \implies \Log z\in R\text{.} \end{align*}
The identities (1.24)–(1.25) are easily verified using the formulas for \(\exp\) and \(\Log\) applied to elements in \(R\) and \(\C-\{0\}\text{.}\)

Example 1.7.5. Logarithm.

Compute \(\log z\) and \(\Log z\) (if applicable) for the given \(z\text{.}\)
  1. \(\displaystyle z=3\)
  2. \(\displaystyle z=-2\)
  3. \(z=1+i\text{.}\)
Solution.
  1. We have
    \begin{align*} \log 3 \amp =\ln 3+i\arg(3)=\{\ln 3+2\pi k i\colon k\in \Z\}\\ \Log 3 \amp =\ln 3\text{.} \end{align*}
  2. We have
    \begin{align*} \log(-2) \amp = \ln\abs{-2}+i\arg(-2)\\ \amp = =\ln 2+i\arg(-2)\\ \amp = \{\ln 2+i(\pi +2\pi k)\colon k\in \Z\}\\ \amp = \{\ln 2+(2k+1)\pi i \colon k\in \Z\}\text{.} \end{align*}
    Technically, \(\Log\) is not defined at \(z=-2\text{.}\)
  3. We have
    \begin{align*} \log(1+i) \amp = \ln(\sqrt{2})+i\arg(1+i)\\ \amp = \{\ln(\sqrt{2})+i(\pi/4+2\pi k)\colon k\in \Z\}\\ \Log(1+i) \amp = \ln(\sqrt{2})+\frac{\pi i}{4}\text{.} \end{align*}

Example 1.7.6. Invalid identity.

Show that \(\Log(z_1z_2)\) is not always equal to \(\Log(z_1)+\Log(z_2)\text{.}\)
Solution.
In general we have
\begin{align*} \Log(z_1z_2) \amp = \ln\abs{z_1z_2}+i\Arg(z_1z_2)\\ \amp =(\ln\abs{z_1}+\ln\abs{z_2})+i\Arg(z_1z_2)\\ \Log z_1+\Log z_2 \amp = (\ln\abs{z_1}+\ln\abs{z_2})+i(\Arg z_1+\Arg z_2)\text{.} \end{align*}
The issue preventing equality to hold in general is that
\begin{equation*} \Arg(z_1z_2)\ne \Arg(z_1)+\Arg(z_2)\text{.} \end{equation*}
Indeed, taking \(z_1=e^{\pi i/2}\) and \(z_2=e^{3\pi i/4}\text{,}\) we have
\begin{align*} \Arg(z_1z_2) \amp = \Arg(e^{5\pi i /4})=-\frac{3\pi}{4}\\ \Arg(z_1)+\Arg(z_2) \amp =\frac{\pi}{2}+\frac{3\pi }{4}=\frac{5\pi}{4}\text{.} \end{align*}

Subsection Complex exponentiation

Definition 1.7.7. Complex exponentiation.

Given \(z\in \C-\{0\}\) and any \(w\in \C\text{,}\) we define the set \(z^w\) as
\begin{equation*} z^w=e^{w\log z}\text{.} \end{equation*}
Given a fixed \(w\in \C\text{,}\) the function \(f\colon \C-(-\infty, 0]\rightarrow \C\) defined as \(f(z)=e^{w\Log z}\) is called the principal branch of the power function with exponent \(w\) (or principal branch of \(f(z)=z^w\)).

Example 1.7.8. Powers.

Compute the following power expressions.
  1. \(\displaystyle i^i\)
  2. \(z^i\text{,}\) \(z\ne 0\text{.}\)
Solution.
  1. We have
    \begin{align*} i^i \amp = e^{i\log(i)}\\ \amp = e^{i(\ln\abs{i}+i\arg(i))}\\ \amp = e^{i(\ln 1+i\arg(i))}\\ \amp = e^{i(0+i\arg(z))}\\ \amp = e^{-\arg(i)}\\ \amp =\{e^{-\pi/2+2\pi k}\colon k\in \Z\}\\ \amp = \{e^{-\pi/2}e^{2\pi k}\colon k\in \Z\}\text{.} \end{align*}
  2. For \(z\ne 0\text{,}\) we have
    \begin{align*} z^i \amp = e^{i\log(z)}\\ \amp = e^{i(\ln\abs{z}+i\arg(z))}\\ \amp = e^{-\arg(z)+i\ln\abs{z}}\\ \amp =\{e^{(-\Arg z+2\pi k)+i\ln\abs{z}}\colon k\in \Z\}\\ \amp = \{e^{-\Arg z+2\pi k}e^{i\ln\abs{z}}\colon k \in \Z\}\\ \amp = \{e^{-\Arg z}e^{2\pi k}e^{i\ln\abs{z}}e^{2\pi k}\colon k \in \Z\}\text{.} \end{align*}

Example 1.7.9. \(n\)-th roots.

Let \(z\ne 0\text{.}\) Show that the set \(z^{1/n}\) is precisely the set of \(n\) distinct \(n\)-th roots of \(z\text{.}\)
Solution.
By definition we have
\begin{align*} z^{1/n} \amp =e^{\frac{1}{n}\log z}\\ \amp = e^{\frac{1}{n}(\ln\abs{z}+i\arg(z))}\\ \amp = \{e^{\frac{1}{n}\ln\abs{z}}e^{i(\Arg(z)/n +2\pi k/n)}\colon k\in \Z\}\\ \amp = \{e^{\ln \abs{z}^{1/n}}e^{i(\Arg(z)/n+2\pi k/n)}\colon k\in \Z\} \\ \amp = \{\abs{z}^{1/n}e^{i(\Arg(z)/n+2\pi k/n)}\colon k\in \Z\}\amp (e^{\ln c}=c)\\ \amp = \{\abs{z}^{1/n}e^{i(\Arg(z)/n+2\pi k/n)}\colon 0\leq k\leq n-1\}\text{,} \end{align*}
which we recognize as the set of all \(n\)-th roots of \(z\text{.}\)
Note that the last equality above follows from the fact that for any \(k'\in \Z\text{,}\) performing division with remainder by \(n\) we can write
\begin{equation*} k'=nq+k \end{equation*}
for some integer \(q\) and remainder \(k\) satisfying \(0\leq k\leq n-1\text{,}\) whence we have
\begin{equation*} 2\pi k'/n=2\pi(k+qn)/n=2\pi k+2\pi q\text{,} \end{equation*}
and thus
\begin{equation*} e^{2\pi i k'/n}=e^{2\pi i k/n}\text{.} \end{equation*}

Subsection Trigonometric functions

Given a real number \(t\in \R\text{,}\) from the identity
\begin{equation*} e^{it}=\cos t+i\sin t \end{equation*}
and our formulas
\begin{align*} \Re z \amp =\frac{z+\overline{z}}{2} \amp \Im z\amp \frac{z-\overline{z}}{2i} \end{align*}
it follows that
\begin{align*} \cos t\amp =\frac{e^{it}+e^{-it}}{2} \amp \sin t\amp =\frac{e^{it}-e^{-it}}{2i} \text{.} \end{align*}
It is then natural to extend these formulas for all complex numbers \(z\text{,}\) yielding our definitions for the complex cosine and sine functions:
\begin{align*} \cos z\amp =\frac{e^{iz}+e^{-iz}}{2} \amp \sin z\amp =\frac{e^{iz}-e^{-iz}}{2i} \end{align*}

Definition 1.7.10. Trigonometric functions.

We define the complex cosine and sine \(\cos\colon \C\rightarrow \C\) and \(\sin\colon \C\rightarrow \C\) as follows:
\begin{align} \cos(z) \amp = \frac{e^{iz}+e^{-iz}}{2} \amp \tag{1.26}\\ \sin(z)\amp =\frac{e^{iz}-e^{-iz}}{2i} \text{.}\tag{1.27} \end{align}
We then further define complex extensions of the other elementary trigonometric functions in the usual way:
\begin{align*} \tan z \amp = \frac{\sin z}{\cos z} \amp \cot z\amp = \frac{\cos z}{\sin z}\\ \sec z \amp = \frac{1}{\cos z} \amp \csc z\amp =\frac{1}{\sin z}\text{.} \end{align*}

Proof.

  1. Given \(z,w\in \C\text{,}\) we have
    \begin{align*} \cos z=\cos w \amp \iff e^{iz}+e^{-iz}=e^{iw}+e^{-iw}\\ \amp \iff e^{iw}e^{iz}+e^{iw}e^{-iz}=e^{iw}e^{iw}+e^{iw}e^{-iw}\\ \amp \iff e^{i(z+w)}+e^{-i(z-w)}=e^{iw+iz}e^{iw-iz}+1\\ \amp \iff e^{i(z+w)}+e^{-i(z-w)}=e^{i(z+w)}e^{-i(z-w)}+1\\ \amp \iff e^{i(z+w)}(1-e^{-i(z-w)})-(1-e^{-i(z-w)})=0\\ \amp \iff (1-e^{-i(z-w)})(e^{i(z+w)}-1)=0 \\ \amp \iff e^{-i(z-w)}=1 \text{ or } e^{i(z+w)}=1\text{.} \end{align*}
    Next since \(e^u=1\) if and only if \(u=0+2\pi k i\) for some \(k\in \Z\text{,}\) we see that \(\cos z=\cos w\) if and only if
    \begin{align*} -i(z-w)\amp =2\pi k i \text{ for some } k\in \Z \\ \amp \text{ or } \\ i(z+w)\amp =2\pi k i \text{ for some } k\in \Z \end{align*}
    if and only if
    \begin{align*} z\amp =w+2\pi k \text{ for some } k\in \Z \\ \amp \text{ or } \\ z\amp=-w +2\pi k \text{ for some } k\in \Z \text{,} \end{align*}
    as claimed.
  2. The proof of this statement is very similar to the one above.
  3. These identities can be proved in a straightforward manner using the the defining formulas (1.26)–(1.27) and properties of complex multiplication.

Definition 1.7.12. Complex hyperbolic functions.

We define the complex hyperbolic functions \(\cosh\colon \C\rightarrow \C\) and \(\sinh\colon \C\rightarrow \C\) as follows:
\begin{align*} \cosh(z) \amp =\frac{e^z+e^{-z}}{2} \amp \sinh(z)=\frac{e^{z}-e^{-z}}{2}\text{.} \end{align*}

Proof.

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