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If a complex integral cannot be evaluated explicitly, then one might like to get an estimate for it. For that purpose, the following lemma is very useful.
### Estimation Lemma
**Lemma 16**.: _Suppose \(f:\gamma^{*}\to\mathbb{C}\) is continuous, where \(\gamma\) is a path with parameter interval \([\alpha,\beta]\) and \(\gamma^{*}=\gamma([\alpha,\beta])\). Then_
\[\left|\int\limits_{\gamma}fdz\right|\leq\int\limits_{\alpha}^{\beta}|f(z)||dz|.\]
_In particular, if there exists a real number \(M\) such that \(|f(z)|\leq M\) for all \(z\in\gamma^{*}\), then_
\[\left|\int\limits_{\gamma}f(z)dz\right|\leq ML,\]
_where \(L\) denotes the length of \(\gamma\), which is, by definition,_
\[L=\int_{\alpha}^{\beta}|\gamma^{\prime}(t)|dt.\]
Note that for circular arcs, line segments and therefore for contours, the definition for \(L\) is exactly what we expect the “length” to be.
Proof
: The idea of the proof follows from the fact that
\[\left|\int_{\gamma}f(z)\,dz\right|\leq\int_{\alpha}^{\beta}|f(\gamma(t)\, \gamma^{\prime}(t)|\ dt.\]
If \(|f(z)|\leq M\) for all \(t\in\gamma^{*}\) then
\[\left|\int_{\gamma}f(z)\,dz\right|\leq M\int_{\alpha}^{\beta}|\gamma^{\prime} (t)|\ dt.\qed\]
### Exercises
1. Define parametrically a path \(\gamma\) for which \(\gamma^{*}\) is: 1. The square with vertices \(1+i\), \(-1+i\), \(-1-i\), \(1-i\). 2. The pair of circles \(|z-1|=1\) and \(|z+1|=1\), one traced clockwise and the other counterclockwise. 3. The closed semicircle in the right half-plane with \(i[-R,R]\) as diameter.
2. Describe the image \(\gamma^{*}\) of \(\gamma\) in the following: 1. \(\gamma(t)=1+ie^{it},\ \ \ t\in[0,\pi]\). 2. \(\gamma(t)=e^{it},\ \ \ t\in[-\pi,2\pi]\).
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### 3.7 Cauchy’s Theorem
#### Cauchy’s Theorem: Basic Version
One form of Cauchy’s theorem states that if \(\gamma\) is a simple closed curve and if \(f\) is holomorphic inside \(\gamma\) and on \(\gamma\) then \(\int_{\gamma}f=0\). This theorem is very central to the theory of holomorphic functions. If the function is not holomorphic on the whole region inside \(\gamma\) then the integral may or may not be zero. For example, \(f(z)=\frac{1}{z}\) on the unit circle \(\gamma\) is holomorphic except at \(z=0\) and
\[\int_{\gamma}\frac{1}{z}\,dz=\int_{0}^{2\pi}\frac{1}{e^{it}}e^{it}idt=2\pi i,\]
while \(f(z)=\frac{1}{z^{2}}\) is still holomorphic except at \(z=0\) and
\[\int_{\gamma}\frac{1}{z^{2}}\,dz=\int_{0}^{2\pi}\frac{1}{e^{2it}}e^{it}idt=0.\]
As stated above, Cauchy’s theorem demonstrates that the path integral of a holomorphic function around a closed curve will be zero provided that the function is also holomorphic on the interior of the curve. To prove the preliminary version of Cauchy’s theorem, we need to recall Green’s Theorem which we state below.
**Theorem 132** (Green’s Theorem).: _Let \(\gamma\) be a simple closed curve oriented counterclockwise in \(\mathbb{C}\) and \(\Omega\) be its interior. If \(P(x,y)\) and \(Q(x,y)\) are two real-valued differentiable functions, then_
\[\int_{\gamma}P(x,y)dx+Q(x,y)dy=\iint_{\Omega}\left(\frac{\partial Q}{\partial x }-\frac{\partial P}{\partial y}\right)\,dx\,dy.\]
For the following theorem recall that by \(H(G)\) we mean the set of holomorphic functions in \(G\).
**Theorem 133** (Preliminary version of Cauchy’s Theorem).: _Let \(G\) be a simply connected region and \(f\in H(G)\) with derivative \(f^{\prime}(z)\) continuous on and inside a simple closed curve \(\gamma\) in \(G\). Then_
\[\int_{\gamma}f(z)dz=0\]
_on every simple closed curve \(\gamma\) in \(G\)._
Proof
: Let \(f(z)=u(z)+iv(z)\), with \(u(z),v(z)\) real-valued functions; we know that Cauchy-Riemann equations hold: \(u_{x}=v_{y}\quad\text{and}\quad u_{y}=-v_{x}\). Then
\[\int_{\gamma}f(z)dz =\int_{\gamma}(u+iv)(dx+idy)\] \[=\int_{\gamma}(udx-vdy)+i\int_{\gamma}(vdx+udy).\] \[=\iint_{\gamma}(-v_{x}-u_{y})dxdy+i\iint_{\gamma}(u_{x}-v_{y})dxdy.\]
Both terms are zero by the Cauchy-Riemann Equations.
**Definition 111**.: We define a **contour** to be a simple closed path whose image is made up of a finite number of line segments and circular arcs. The positive direction on a contour \(\gamma\) is the direction corresponding to increasing values of the parameter \(t\). The positive direction roughly corresponds to the direction that a person must walk on \(\gamma\) in order to keep the interior of \(\gamma\) to the left. For example, the circle \(z(t)=e^{it}\) for \(0\leq t\leq 2\pi\) has a positive orientation. The negative direction on a contour \(\gamma\) is the direction opposite to the positive direction. If \(\gamma\) has an orientation, the opposite curve, that is a curve with opposite orientation, is denoted by \(-\gamma\). The purpose of introducing contours is to bypass the geometric complexity one encounters working with general closed paths. In particular we will assume the following restricted form of the well-known Jordan curve theorem which asserts that a simple closed path has an “inside” and an “outside.” Suppose \(\gamma\) is a curve with parameter interval \([\alpha,\beta]\), then by \(\gamma^{*}\) we mean the image set of \(\gamma\), in other words \(\gamma^{*}:=\{\gamma(t):\ t\in[\alpha,\beta]\}\).
**Theorem 134** (Jordan curve theorem for a contour).: _Let \(\gamma\) be a contour. Suppose that the inside of \(\gamma\) is denoted by \(I(\gamma)\) and the outside of \(\gamma\) is denoted by \(O(\gamma)\). Then the complement of \(\gamma^{*}\) is of the form_
\[I(\gamma)\cup O(\gamma)\quad\text{ with }\quad I(\gamma)\cap O(\gamma)=\emptyset.\]
_Moreover both are open, connected sets with \(I(\gamma)\) bounded and \(O(\gamma)\) unbounded._
For proof of Jordan’s curve theorem, consult books on algebraic topology, such as [25].
**Remark 74**.: Cauchy’s theorem is often proved for a triangle first. For this version, suppose that \(f\) is holomorphic inside and on a triangle \(\gamma\). Then \(\int_{\gamma}f(z)\,dz=0.\) There is also topologically more sophisticated approaches to Cauchy’s theorem. For example, Cauchy’s theorem for a contour assumes that \(f\) is holomorphic inside and on a contour \(\gamma\), then \(\int_{\gamma}f(z)\,dz=0.\)
### Deformation Theorem
It is important to be able to study functions which are not holomorphic on the entire \(\gamma\) and whose integral might not be zero. For example, \(\frac{1}{z}\) fails to be holomorphic at \(z=0\) and
\[\int_{\gamma}f(z)dz=2\pi i,\]
where \(\gamma\) is the unit circle (the point \(z=0\) is called the singularity of \(f\)). To study such functions, we replace \(\int_{\gamma}f(z)dz\) by \(\int_{\widetilde{\gamma}}f(z)dz\) where \(\widetilde{\gamma}\) is a simpler curve (say, a circle) where \(\int_{\widetilde{\gamma}}f(z)dz\) can often be evaluated. The procedure that allows us to pass from \(\gamma\) to \(\widetilde{\gamma}\) is based on Cauchy’s Theorem and is as follows. We first need the following definition.
### Preliminary Version of the Deformation Theorem
Let \(f\) be holomorphic in a region \(G\) and let \(\gamma\) be a simple closed curve in \(G\). Suppose that \(\gamma\) can be continuously deformed to another simply closed curved \(\widetilde{\gamma}\) without passing outside of \(G\). (In such a case, we say that \(\gamma\) is _homotopic_ to \(\widetilde{\gamma}\) in \(G\)). Then
\[\int_{\gamma}f(z)dz=\int_{\widetilde{\gamma}}f(z)\,dz. \tag{3.2} \]\]
Another version of the theorem is as follows: if \(\gamma\) is a positively oriented contour, \(\overline{D}(a;r)\subset I(\gamma)\) and \(f\) is holomorphic inside \(\gamma\) except possibly at \(a\), then
\[\int_{\gamma}f(z)dz=\int_{\gamma(a;r)}f(z)\,dz.\]
Note that with this version we can replace an integral around a contour \(\gamma\) by an integral around a circle, centered at a, inside \(\gamma\).
Note that \(f\) need not be analytic inside \(\gamma\), so that Cauchy’s Theorem does not imply that the integrals in equation (3.2) are zero. It is implicit in the statement of the deformation theorem that both \(\gamma\) and \(\widetilde{\gamma}\) are oriented in the counterclockwise direction (Figure 3.31).
Proof
: The strategy is to form a closed path to which Cauchy’s theorem can be applied. Consider the following figure, in which a curve \(\gamma_{0}\) is drawn joining \(\gamma\) to \(\widetilde{\gamma}\). We assume that such a curve can in fact be drawn (Figure 3.32).
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