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### 3.4 Complex Logarithmic Function
The polar representation of a complex number \(z=|z|e^{i\theta}\) is not uniquely determined. This is the main reason for multi-valuedness in complex analysis. For \(z\neq 0\), we define the argument of \(z\) to be the set of values \(\{\theta\in\mathbb{R}:z=|z|e^{i\theta}\}\). In fact, when we say argument of \(z\), we mean not a single number, but an infinite set. More precisely we can define \(\operatorname{Arg}z\) (note the capital letter) as the set of values
\[\operatorname{Arg}z=\{\theta+2k\pi\ \ \text{ for }\ k\in\mathbb{Z}\}.\]
Figure 3.17: Image of a horizontal line under \(w=e^{z}\)
For example,
\[\mathrm{Arg}\,i =\frac{\pi}{2}+2k\pi\] \[=(4k+1)\frac{\pi}{2}\enspace,\,k\in\mathbb{Z}.\]
If we restrict \(\theta\) to \((-\pi,\pi]\), we can have a single value for \(\mathrm{Arg}\,z\); we call this the _principal value_ of the argument and use the notation \(\mathrm{arg}\,\,z\). It should be emphasized that the choice of the interval \((-\pi,\pi]\) is completely arbitrary: we might have chosen the interval \([0,2\pi)\) instead.
In real analysis, the statements \(y=e^{x}\) and \(x=\log y\) where \(y>0\) are equivalent. If we try to use this approach to define \(\log\,\,z\) for a complex \(z\), we encounter a difficulty. To define complex logarithms as an inverse of an exponential requires some care, since \(e^{w}=z\) has infinitely many solutions. Let us suppose that \(w=\log z\) is equivalent to \(z=e^{w}\), where \(w=u+iv\). Then
\[z=e^{w}=e^{u+iv}=e^{u}\cdot e^{iv}\]
and so
* \(e^{u}=|z|\). Thus \(u=\log|z|\) (we are using a simple \(\log\) to denote \(\log_{e}\)).
* \(v=\mathrm{argument\,\,of}\,z\).
Thus \(e^{w}=z\) will yield \(w=\log|z|+i\mathrm{Arg}\,z\). However we just noted that there are infinitely many \(\mathrm{Arg}z\), thus when we write
\[\mathrm{Log}z=\{\log|z|+i\theta:\ \theta\in\mathrm{Arg}(z)\}\]
we again mean a multi-valued function and restricting \(\theta\) to either \([0,2\pi)\) or \((-\pi,\pi]\) gives a single value which we call the _principle value or principal branch of the logarithmic function_ expressed as
\[\log z=\log|z|+i\arg z.\]
It follows that the choice of principal logarithm is arbitrary; however, once we make the choice of \((-\pi,\,\pi]\), we shall stick to it. When we restrict \(\theta\) to \((-\pi,\pi]\), we basically are crossing out the negative \(x\)-axis. We think of the plane as having a cut along the negative \(x\)-axis, preventing the argument of \(z\) from leaving the interval \((-\pi,\,\pi]\) (Figure 3.18).
In the definition of the principal argument and the principal logarithm, the position of the cut is arbitrary. The positive \(x\)-axis would do as well so would any half line containing the origin. The key issue here is that the cut should contain \(0\) and should go off the infinity. We say \(0\) and \(\infty\) are _branch points_. If \(z\) moves around any closed path contained in \(\mathbb{C}\setminus(-\infty,0]\), the logarithm changes continuously and returns to its original value.
**Proposition 41**.: _Let \(D^{*}=\mathbb{C}\setminus(-\infty,0]\). Then_
1. \(\log z\) _is continuous on_ \(D^{*}\)_._
2. \(\log z\) _is holomorphic on_ \(D^{*}\) _and if_ \(f_{1}(z)=\) _principal value of_ \(\log z\)_, then_ \(f_{1}^{\prime}(z)=\dfrac{1}{z}\)_._
Proof
:
1. Notice that when we take the principal value of \[\log z =u(x,y)+iv(x,y)\] \[=\log|z|+i\arg z,\] \(u(x,y)=\log|z|=\log\sqrt{x^{2}+y^{2}}\) is continuous everywhere except at \((0,0)\), and \(v(x,y)=\arg z\) is discontinuous at every point on the negative \(x\)-axis.
2. Let \(z=re^{i\theta}\)\(-\pi<\theta\leq\pi\). Then \[\log z=\log r+i\theta=u(r,\theta)+iv(r,\theta),\] where \(u(r,\theta)=\log r\) and \(v(r,\theta)=\theta\). Taking partial derivatives, we have \[\dfrac{\partial u}{\partial r}=\dfrac{1}{r},\qquad\dfrac{\partial u}{\partial \theta}=0,\qquad\dfrac{\partial v}{\partial r}=0\quad\text{and}\quad\dfrac{ \partial v}{\partial\theta}=1.\]
Figure 3.18: \(\mathbb{C}\setminus(-\infty,0]\): branch points and the cut
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**Example 100** (Riemann surface for \(y=e^{z}\)).: The theory of Riemann surfaces provide a framework for a study of multi-functions such as \(w=e^{z}\), \(\log\,z\) or \(\sqrt{z}\). The main idea is instead of working with individual branches of a multi-function in a cut plane, consider all branches as a single function on a domain set consisting of many copies of the plane. Since we will glue these copies together, one can pass continuously from one branch of the multi-function to another. To clarify, let us consider a Riemann surface for \(w=e^{z}\) on the plane \(x\leq 0\). Define the fundamental region as an infinite strip \(S_{n}\) where
\[S_{n}=\{(x,y):\ x\leq 0,\quad(2n-1)\pi for \(n=0,\pm 1,\pm 2,\dots\) Notice that \(w=e^{z}\) maps each infinite strip \(S_{n}\) onto the punctured unit disk \(D_{n}=0<|w|\leq 1\) shown in Figure 3.19. In other words, \(e^{z}\) describes an infinite-to-one covering of the punctured disk \(D_{n}\). To visualize this covering, imagine that there is a different image disk \(D_{n}\) for each half infinite strip \(S_{n}\). Now cut each disk \(D_{n}\) open along \(-1\leq u<0\). To form the Riemann surface for \(w=e^{z}\) attach for each \(n\), the cut disk \(D_{n}\) to the cut disk \(D_{n+1}\) along the edge. Think of this surface in \(xyz\)-space so that for each \(z\) in the half-plane, the images \(\dots z_{-1},z_{0},z_{1},\dots\) of \(z\) in \(\dots D_{-1},D_{0},D_{1},\dots\), respectively lie directly above the point \(w=e^{z}\) in the \(xy\)-plane. In other words if one projects the points on the Riemann surface vertically down onto the \(xy\)-plane one can see the infinite-to-one nature of the mapping \(w=e^{z}\) (Figure 3.20). Readers interested in pursuing the study of Riemann surfaces will have no difficulty finding relevant literature, e.g., [47]. ### Trigonometric and Hyperbolic Functions For \(z\in\mathbb{C}\), we define: \[\cos z:=\sum_{n=0}^{\infty}(-1)^{n}\frac{z^{2n}}{(2n)!}\ ;\ \sin z:=\sum_{n=0}^{ \infty}(-1)^{n}\frac{z^{2n+1}}{(2n+1)!}\ \ \ \mbox{and}\] \[\cosh z:=\sum_{n=0}^{\infty}\frac{z^{2n}}{(2n)!}\ ;\ \sinh z:=\sum_{n=0}^{ \infty}\frac{z^{2n+1}}{(2n+1)!}.\] Since all of these four functions are holomorphic in \(\mathbb{C}\), we can find their derivatives differentiating term by term, and it can be shown that \[\frac{d}{dz}(\cos z)=-\sin z\,\ \frac{d}{dz}(\sin z)=\cos z,\] \[\frac{d}{dz}(\cosh z)=\sinh z\,\ \frac{d}{dz}(\sinh z)=\cosh z.\] From \(e^{iz}=\cos z+i\sin z\), we also have \[\cos z=\frac{e^{iz}+e^{-iz}}{2}\ \mbox{and}\ \sin z=\frac{e^{iz}-e^{-iz}}{2i},\] similarly \[\cosh z=\frac{e^{z}+e^{-z}}{2}\ \mbox{and}\ \sinh z=\frac{e^{z}-e^{-z}}{2}.\] **Proposition 42**.: _We have the following equalities:_ * \(\sin^{2}z+\cos^{2}z=1\) Figure 3.20: Riemann surface for \(w=e^{z}\)_._ : This follows directly from the definitions. For example \[\sin^{2}z+\cos^{2}z =\left(\frac{e^{iz}-e^{-iz}}{2i}\right)^{2}+\left(\frac{e^{iz}+e^{ -iz}}{2}\right)^{2}\] \[=\frac{e^{2iz}-2+e^{-2iz}}{4}+\frac{e^{2iz}+2+e^{-2iz}}{4}=1.\] We leave parts b) and c) to the reader to verify. It is also quite straightforward to see that the complex trigonometric functions are periodic. Since if \(f\) is holomorphic on some open set \(G\), then \(1/f\) is holomorphic in \(G\) provided \(f(z)\neq 0\) in \(G\). Thus, it is good to locate zeros of the trigonometric functions. It turns out that \[\cos z=0\iff z=\frac{1}{2}(2k+1)\pi\text{, }k\in\mathbb{Z}\] \[\sin z=0\iff z=k\pi\text{, }k\in\mathbb{Z}\] and \[\cosh z=0\iff z=\frac{1}{2}(2k+1)\pi i\text{, }k\in\mathbb{Z}\] \[\sinh z=0\iff z=k\pi i\text{, }k\in\mathbb{Z}.\] **Remark** For \(x\in\mathbb{R}\), \(|\sin x|\leq 1\) and \(|\cos x|\leq 1\) are quite useful inequalities. For \(z\in\mathbb{C}\) these inequalities are no longer valid; in fact both functions \(\sin z\) and \(\cos z\) are unbounded in \(\mathbb{C}\). See Exercise 4 at the end of this section.
2. \(\sin(z+w)=\sin z\,\cos w+\cos z\,\sin w\)_._
3. \(\cos(z+w)=\cos z\,\cos w-\sin z\,\sin w\)_._Proof
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