Convention 1.14.1. Paths.
From now on, unless explicitly stated otherwise, a path \(\gamma\colon [a,b]\rightarrow \C\) will be assumed to be piecewise smooth.
Section 1.14 Cauchy-Goursat theorem
In this section we meet the famous Cauchy-Goursat theorem: our first truly deep theorem, and one that is fundamental to all the theoretical highlights that follow.
Subsection Existence of antiderivatives
As we observed in Corollary 1.13.8, if \(f\) has an antiderivative on the open set \(U\text{,}\) then \(\int_\gamma f\, dz=0\) for any closed path lying in \(U\text{.}\) Equivalently, if \(\int_\gamma f\, dz\ne 0\) for some closed path lying in \(U\text{,}\) then there is no antiderivative of \(f\) on \(U\text{.}\)
Applying this result to \(f(z)=1/z\text{,}\) we see that for any open connected set \(U\) containing \(0\text{,}\) there is no antiderivative of \(f\) on \(U-\{0\}\text{:}\) indeed, the path \(\gamma(t)=re^{it}\) lies in \(U\) for \(r\) small enough, and we have computed that \(\int_\gamma 1/z\, dz=2\pi i\ne 0\) for any such path. As a further consequence, it follows that there can be no branch of \(\log\) on an open set of the form \(U-\{0\}\) where \(U\) contains \(0\text{,}\) since as we have seen previously, any such branch would be an antiderivative of \(f\text{!}\)
The example illustrates the subtlety of the question of whether an antiderivative exists for a continuous complex function. The next theorem tells us that the implication from Corollary 1.13.8 is in fact an equivalence if \(U\) is open and connected.
Theorem 1.14.2. Antiderivative theorem.
Let \(U\subseteq \C\) be open and connected, and assume \(f\) is continuous on \(U\text{.}\) The following statements are equivalent.
-
Existence of antiderivative.\(f\) has an antiderivative on \(U\text{.}\)
-
Closed paths.\(\int_\gamma f\, dz=0\) for any closed path \(\gamma\) lying in \(U\text{.}\)
-
Independence of paths.\(\int_\alpha f\, dz=\int_\beta f\, dz\) for any paths \(\alpha\) and \(\beta\) lying in \(U\) that have the same initial and terminal points.
Proof.
Implication: (1)\(\implies\)(2).
Let \(F\) be an antiderivative of \(f\) on \(U\text{.}\) Given a closed piecewise smooth path \(\gamma\text{,}\) let \(z_0=z_1\) be its initial/terminal point. By the fundamental theorem of calculus for line integrals, we have
\begin{equation*}
\int_\gamma f\, dz=F(z_1)-F(z_0)=F(z_0)-F(z_0)=0\text{.}
\end{equation*}
Implication: (2)\(\implies\)(3).
Since \(\alpha\) and \(\beta\) share the same initial and terminal points, the curve \(\gamma=\alpha* (-\beta)\) is closed, and we have
\begin{equation*}
0=\int_\gamma f\, dz=\int_\alpha f\, dz+\int_{-\beta}f\, dz=\int_\alpha f\, dz-\int_\beta f\, dz\text{.}
\end{equation*}
It follow that
\begin{equation*}
\int_\alpha f\, dz=\int_\beta f\, dz\text{,}
\end{equation*}
as desired.
Implication: (3)\(\implies\)(1).
Pick any element \(z^*\in U\text{.}\) Since \(U\) is open and connected, it is polygonally connected. Thus for any \(z\in U\) there exists a polygonal path beginning at \(z^*\) and ending at \(z\text{.}\) We define the function \(F\colon U\rightarrow \C\) as
\begin{equation}
F(z)=\int_{z^*}^z f\, dz\text{,}\tag{1.54}
\end{equation}
where \(\int_{z^*}^z f\, dz\) denotes the contour integral along any polygonal path from \(z^*\) to \(z\text{.}\) Since we assume that line integrals of \(f\) are path independent, our function \(F\) is well-defined: i.e., \(F(z)\) does indeed return a single well-defined value, no matter which path you choose. It remains to show that \(F\) is an antiderivative of \(f\) on \(U\text{.}\)
Pick any \(z_0\in U\text{.}\) Firstly, since \(U\) is open we can find a \(\delta> 0\) such that \(B_{\delta}(z_0)\subseteq U\text{.}\) We will prove that \(F'(z_0)\) by showing that
\begin{equation*}
F(z)-F(z_0)=f(z_0)(z-z_0)+r(z)
\end{equation*}
for all \(z\in B_{\delta}(z_0)\text{,}\) where \(r(z)\) satisfies
\begin{equation*}
\lim\limits_{z\to z_0}\frac{\abs{r(z)}}{\abs{z-z_0}}=0\text{.}
\end{equation*}
To this end note that for all \(z\in B_{\delta}(z_0)\text{,}\) we have
\begin{align*}
F(z)-F(z_0) \amp = \int_{z^*}^zf\, dz-\int_{z^*}^{z_0} f\, dz\\
\amp = \int_{z^*}^{z_0}f\, dz +\int_{\gamma_{z_0,z}}f\, dz-\int_{z^*}^{z_0}f\, dz\\
\amp = \int_{\gamma_{z_0,z}}f\, dz \text{,}
\end{align*}
where \(\gamma_{z0,z1}\) is the straight line path from \(z_0\) to \(z\text{.}\) Next, we have
\begin{align*}
\int_{\gamma_{z_0,z}}f\, dz \amp = \int_{\gamma_{z_0,z}}f(z_0)+f(z)-f(z_0)\, dz\\
\amp = \int_{\gamma_{z_0,z}}f(z_0)\, dz+\int_{\gamma_{z_0,z}}f(z)-f(z_0)\, dz\\
\amp =f(z_0)(z-z_0)+\int_{\gamma_{z_0,z}}f(z)-f(z_0)\, dz\text{,}
\end{align*}
where the last step follows from the fact that \(G(z)=z\) is an antiderivative of \(1\text{,}\) and thus
\begin{equation*}
\int_{\gamma_{z_0,z}}f(z_0)\, dz=f(z_0)\int_{\gamma_{z_0,z}}1\, dz=f(z_0)(z-z_0)\text{,}
\end{equation*}
by the fundamental theorem of calculus for line integrals.
Letting \(r(z)=\int_{z_0,z}f(z)-f(z_0)\, dz\text{,}\) we then have
\begin{equation*}
F(z)-F(z_0)=f(z_0)(z-z_0)+r(z)\text{.}
\end{equation*}
It remains to show that \(\lim\limits_{z\to z_0}\abs{r(z)}/{\abs{z-z_0}}=0\text{.}\) To this end, given any \(\epsilon > 0\text{,}\) shrink \(\delta\) as necessary so that \(\abs{z-z_0}< \delta\) implies \(\abs{f(z)-f(z)}< \epsilon\text{.}\) We then have
\begin{align*}
\frac{\abs{r(z)}}{\abs{z-z_0}} \amp = \frac{\abs{\int_{z_0,z}f\, dz}}{\abs{z-z_0}} \\
\amp \leq \frac{\epsilon \abs{z-z_0}}{\abs{z-z_0}} \amp (ML-\text{ineq.})\\
\amp \leq \epsilon\text{.}
\end{align*}
Since \(\epsilon\) was abribitrary, we see that \(\lim\limits_{z\to z_0}\abs{r(z)}/\abs{z-z_0}=0\text{,}\) as desired.
Subsection Cauchy-Goursat theorems
Convention 1.14.3. Triangles.
Given three non-colinear complex numbers \(z_0,z_1,z_2\text{,}\) we denote by \(\gamma_{(z_0,z_1,z_2)}\) any simple closed path of the form
\begin{equation*}
\gamma_{(z_1,z_2)}*\gamma_{(z_2,z_3)}*\gamma_{(z_3,z_0)}\text{,}
\end{equation*}
where \(\gamma_{(z_i,z_j)}\) denotes a simple straight-line path from \(z_i\) to \(z_k\text{.}\) We denote by \(\triangle(z_0,z_1,z_2)\) the triangular region consisting of \(\gamma_{(z_1,z_2,z_3)}\) and all points lying within it. It is easy to see that \(\triangle=\triangle(z_0,z_1,z_2)\) is closed and that \(\partial \triangle=\gamma_{(z_1,z_2,z_3)}\text{.}\)
Theorem 1.14.4. Cauchy-Goursat: triangles.
Assume \(f\) is holomorphic on \(U\text{.}\) Given any triangle \(\triangle z_0,z_1,z_2\) contained in \(U\text{,}\) we have
\begin{equation*}
\int_{\gamma_{(z_0,z_1,z_2)}}f\, dz=0\text{.}
\end{equation*}
Proof.
We give a near complete proof, save for one unproved fact about closed and bounded sets (aka., compact sets). Let \(\Delta=\Delta(z_0,z_1,z_2)\) and let \(\gamma=\gamma_{(z_0,z_1,z_2)}\text{.}\)
Idea: compare \(\int_\gamma f\, dz\) with integral over smaller triangles.
Let \(w_0,w_1,w_2\) be the midpoints of the sides \(\Delta\) and consider the four closed paths
\begin{align*}
\gamma_1 \amp = \gamma_{z_0,w_0,w_1}\\
\gamma_2 \amp = \gamma_{w_0,z_1,w_1}\\
\gamma_3 \amp = \gamma_{w_1,z_2,w_2}\\
\gamma_4 \amp = \gamma_{w_2,w_0,w_1}\text{.}
\end{align*}
See Figure 1.14.5. Using elementary integral properties, thanks to the cancellation that occurs due to \(\gamma_4\) we have
\begin{equation*}
\int_\gamma f\, dz=\int_{\gamma_1} f\, dz+\int_{\gamma_2} f\, dz+\int_{\gamma_3} f\, dz+\int_{\gamma_4} f\, dz\text{.}
\end{equation*}
Picking any \(1\leq k_0\leq 4\) satisfying
\begin{equation*}
\max\left\{\abs{\int_{\gamma_k} f\, dz}\colon 1\leq k\leq 4\right\}=\abs{\int_{\gamma_{k_0}}f\, dz} \text{,}
\end{equation*}
we then have
\begin{align*}
\abs{\int_\gamma f\, dz} \amp \leq \sum_{k=1}^4\abs{\int_{\gamma_k}f\, dz}\\
\amp \leq 4 \abs{\int_{\gamma_{k_0}} f\, dz}\text{.}
\end{align*}
Define \(\gamma^{(1)}=\gamma_{k_0}\text{,}\) and let \(\Delta^{(1)}\) be triangle enclosed by \(\gamma^{(1)}\text{.}\) Iterating this procedure on \(\Delta^{(1)}\text{,}\) we get a sequence of nested triangles
\begin{equation}
\Delta\supseteq \Delta^{(1)}\supseteq \Delta^{(2)}\supseteq \dots\tag{1.55}
\end{equation}
and whose corresponding closed paths \(\gamma^{(n)}\) satisfy (by induction)
\begin{align*}
\abs{\int_\gamma f\, dz} \amp \leq 4^n\abs{\int_{\gamma^{(n)}}f\, dz}\\
L(\gamma^{(n)}) \amp =\frac{L(\gamma)}{2^n}\text{,}
\end{align*}
where here \(L(\phi)\) denotes the arc length of any piecewise smooth path \(\phi\text{.}\)
Pick point in \(\bigcap\limits_{n=1}^\infty \Delta^{(n)}\).
We now use what really is a result about compact sets. We will refrain from giving the general topological definition of compactness; suffice it to say that in the context of \(\C\text{,}\) a compact set is just a closed and bounded set. A general fact about compact sets is that for any descending nested sequence of compact subsets like the one in (1.55), the intersection of the nested sets is nonempty. Thus we can find a \(z_0\in \bigcap_{n=1}^\infty \Delta^{(n)}\) that lies within all of our nested triangles. The remainder of the proof will try and give useful bounds on \(\abs{\int_{\gamma^{(n)}}f\, dz}\) by using the fact that \(z_0\in \Delta^{(n)}\) and \(f\) is differentiable at \(z_0\text{.}\)
Estimating \(\abs{\int_{\gamma^{(n)}}f\, dz}\) using \(f'(z_0)\).
Since \(f\) is differentiable at \(z_0\text{,}\) we have
\begin{equation*}
f(z)=f(z_0)+f'(z_0)(z-z_0)+r(z)
\end{equation*}
where
\begin{equation}
r(z)=f(z)-f(z_0)-f'(z_0)(z-z_0)\tag{1.56}
\end{equation}
is continuous on \(\Delta\) and satisfies
\begin{equation}
\lim\limits_{z\to z_0}\frac{\abs{r(z)}}{\abs{z-z_0}}=0.\tag{1.57}
\end{equation}
Note that the function
\begin{equation*}
g(z)=f(z_0)+f'(z_0)(z-z_0)
\end{equation*}
is a linear polynomial with antiderivative \(G(z)=f(z_0)z+\frac{1}{2}f'(z_0)(z-z_0)^2\text{.}\) It follows from Corollary 1.13.8 that \(\int_{\gamma^{(n)}}g(z)\, dz=0\text{,}\) and thus
\begin{align*}
\int_{\gamma^{(n)}}f\, dz \amp = \int_{\gamma^{(n)}}g(z)+r(z)\, dz\\
\amp = \int_{\gamma^{(n)}}g(z)\, dz+\int_{\gamma^{(n)}}r(z)\, dz\\
\amp =\int_{\gamma^{(n)}} r(z)\, dz \text{.}
\end{align*}
Proof that \(\int_{\gamma}f\, dz=0\).
To reach our desired conclusion, we will show that
\begin{equation*}
\abs{\int_{\gamma}f\, dz}\leq \epsilon
\end{equation*}
for any \(\epsilon > 0\text{.}\) To this end, fix \(\epsilon > 0\text{.}\) Since the function \(r(z)\) defined above satisfies the inequality (1.57), for \(z\) sufficiently close to \(z_0\) we have
\begin{equation*}
\abs{r(z)} < \abs{z-z_0}\frac{\epsilon}{L(\gamma)^2} \text{.}
\end{equation*}
Clearly, we can pick \(n\) large enough so that this inequality holds for all \(z\in \Delta^{(n)}\text{.}\) A straightforward geometric argument shows that \(\abs{z-z_0}\leq L(\gamma^{(n)})\) for all \(z\in \Delta^{(n)}\text{,}\) and thus that
\begin{equation*}
\abs{r(z)}< \frac{L(\gamma^{(n)})\epsilon}{L(\gamma)^2}
\end{equation*}
for all \(z\in \Delta^{(n)}\text{.}\) We conclude that
\begin{align*}
\abs{\int_\gamma f\, dz} \amp \leq 4^n\abs{\int_{\gamma^{(n)}}f\, dz}\\
\amp =4^n\abs{\int_{\gamma^{(n)}}r(z)\, dz} \\
\amp \leq 4^n\cdot \max\{r(z)\colon z\in \Delta^{(n)}\}\cdot L(\gamma^{(n)}) \amp (ML-\text{ineq.})\\
\amp < 4^n \cdot \frac{L(\gamma^{(n)})\epsilon}{L(\gamma)^2} \cdot L(\gamma^{(n)}) \\
\amp = 4^n \cdot \frac{L(\gamma)\epsilon}{2^n L(\gamma)^2} \cdot \frac{L(\gamma)}{2^n}\\
\amp =\epsilon\text{,}
\end{align*}
as desired.
Definition 1.14.6. Star-shaped sets.
A set \(D\subseteq \C\) is star shaped if there is an element \(z^*\) such that for all \(z\in D\) the line segment between \(z\) and \(z^*\) lies in \(D\text{.}\)
Remark 1.14.7. Star-shaped regions.
Observe that since a star-shaped region is clearly polygonally connected, it is connected.
Theorem 1.14.8. Cauchy-Goursat: star-shaped regions.
Assume \(U\subseteq \C\) is open and star-shaped.
-
Closed paths.If \(f\) is holomorphic on \(U\text{,}\) then \(\int_\gamma f\, dz=0\) for any closed path \(\gamma\) lying in \(U\text{.}\)
-
Antiderivatives.If \(f\) is holomorphic on \(U\text{,}\) then \(f\) has an antiderivative on \(U\text{.}\)
Proof.
The two statements are equivalent by Theorem 1.14.2. Thus it suffices to prove (2). We will proceed in a manner much like the proof of Theorem 1.14.2. Pick a star-center \(z^*\) of \(U\text{.}\) For any \(z\in U\) we define \(F(z)=\int_{\gamma_{z^*,z}}f\, dz\text{,}\) where for any two points \(w, w'\) we denote by \(\gamma_{w,w'}\) the straight-line path between them. Note that the question of \(F\) being well defined does not even arise here, as we have specified an explicit path to compute the integral over.
As in the proof of Theorem 1.14.2, we first pick a \(\delta\) such that \(B_{\delta}(z_0)\subseteq U\text{.}\) As long as we can show that
\begin{equation}
F(z)-F(z_0)=\int_{\gamma_{z_0,z}}f\, dz\tag{1.58}
\end{equation}
for all \(z\in B_{\delta}(z_0)\) the rest of the proof of Theorem 1.14.2 follows to show that \(F'(z_0)=f(z_0)\text{.}\) Showing that (1.58) holds in this situation requires a little work, however, since we cannot assume that line integrals are path independent! Instead, we make use of Theorem 1.14.4. Using the fact that \(U\) is star-shaped, it is easy to see that for any \(z\in B_{\delta}(z_0)\text{,}\) the entire triangle \(\Delta(z^*,z_0,z)\) is contained in \(U\text{.}\) The Cauchy-Goursat theorem for triangles then applies, yielding
\begin{equation*}
0=\int_{\gamma_{z^*,z_0,z}}f\, dz=\int_{\gamma_{z^*,z_0}}f\, dz+\int_{\gamma_{z_0,z}}f\, dz+\int_{\gamma_{z,z_0}}f\, dz\text{,}
\end{equation*}
or equivalently,
\begin{equation*}
\int_{\gamma_{z_0,z}}f\, dz=\int_{\gamma_{z^*,z}}f\, dz-\int_{\gamma_{z^*,z_0}}f\, dz=F(z)-F(z_0)\text{,}
\end{equation*}
as desired.
Definition 1.14.9. Elementary regions.
An open connected set \(U\) is elementary if
\begin{equation*}
\int_\gamma f\, dz=0
\end{equation*}
for all paths \(\gamma\) lying in \(U\) and all functions \(f\) that are holomorphic on \(U\text{.}\) Equivalently, an open connected set \(U\) is elementary if every holomorphic function on \(U\) has an antiderivative on \(U\text{.}\)
Remark 1.14.10. Elementary regions.
According to Theorem 1.14.8, any open star-shaped region is elementary. In particular, open balls are elementary.
Corollary 1.14.11. Existence of log branches.
Assume \(U\) is elementary. If \(f\) and \(f'\) are holomorphic on \(U\) and \(f(z)\ne 0\) for all \(z\in U\text{,}\) then there exists a holomorphic branch of \(\mathcal{F}=\log (f(z))\) on \(U\text{.}\)
Proof.
Left as an exercise.
Theorem 1.14.12. Elementary region building.
If \(U_1\) and \(U_2\) are elementary regions, and if \(U_1\cap U_2\) is connected, then \(U_1\cup U_2\) is elementary.
Proof.
Left as an exercise.
Example 1.14.13. Star-shaped and elementary regions.
It is not difficult to show that the following regions (and thus any transformations thereof by a rigid motion) are star-shaped:
- \(\C-R_{\alpha}\text{,}\) \(\alpha\in \R\text{,}\)
- \(\{z\in \C\colon \abs{z}> r, \pi/2-\alpha< \Arg_{0} z< \pi/2+\alpha\}\text{,}\) \(r> 0\text{,}\) \(0< \alpha < \pi/2\text{.}\)
- \(\{z\in \C\colon r < \abs{z}< R, \pi/2-\alpha< \Arg_{0} z< \pi/2+\alpha\}\text{,}\) \(0< r < R\text{,}\) \(0< \alpha < \pi/2\text{.}\)
It follows from Theorem 1.14.12 the following region types (and any transformation thereof by a rigid motion) are elementary:
- \(\{z\in \C\colon r < \abs{z}< R, \alpha< \Arg_{0} z< \beta\}\text{,}\) \(0< r < R\text{,}\) \(0< \alpha < \beta\leq 2\pi\text{,}\)
- \(\{z\in \C\colon r < \abs{z}< R, \Arg_0 z\ne \alpha\}\text{,}\) \(0< r < R\text{,}\) \(0< \alpha \leq 2\pi\text{.}\)
Remark 1.14.16. Elementary versus simply connected.
Starting with some simple star-shaped regions like open discs, we can use Theorem 1.14.12 to build an extensive collection of elementary domains. Indeed for any explicit example we consider, these tools will suffice to show that a given region is in fact elementary.
That said, it is desirable to have some geometric intuition about whether an open connected set is elementary. As it turns out, there is a purely topological characterization of which subsets of \(\C\) are elementary: namely, \(U\subseteq \C\) is elementary if and only if it is open and simply connected. The notion of a simply connected set is a tad too technical for us to include in this course, but the intuitive idea is that an open connected set \(U\) is simply connected if given any two paths \(\alpha\) and \(\beta\) in \(U\) with the same endpoints, we can be continuously deform \(\alpha\) to \(\beta\) without leaving \(U\text{.}\) Another, even less technical intuitive description of a simply connected region is that it is connected and contains no holes.
This intuitive notion of simply connected regions (and its relation to elementary regions) will suffice for our purposes. In more detail, given an open connected set \(U\text{,}\) we can use this intuition to identify it informally as simply connected or not, and hence as elementary or not. If it is then required of us to prove a given region actually is elementary, then we will be able to do so using Theorem 1.14.12.
In the spirit of Remark 1.14.16, we state one more version of the Cauchy-Goursat theorem for connected regions bounded by piecewise smooth closed curves. Significant work must be done to show this last version follows from our previous versions. In particular, the result relies on the famous Jordan curve theorem, which says that if \(\phi\) is a closed piecewise smooth and simple parametrization of the curve \(\mathcal{C}\) the region within \(\mathcal{C}\) is simply connected.
Theorem 1.14.18. Cauchy-Goursat: boundary curve.
Let \(U\) be an open and connected set whose boundary curve \(\partial U\) is the disjoint union of closed piecwise smooth and simple paths \(\gamma_k\text{,}\) \(1\leq k\leq n\text{,}\) and suppose further that each path \(\phi_k\) is positively oriented with respect to \(U\text{.}\) If \(f\) is holomorphic on \(U\) and \(\partial U\text{,}\) then
\begin{equation*}
\int_{\partial U} f\, dz=\sum_{k=1}^{n}\int_{\gamma_k}f\, dz=0\text{.}
\end{equation*}
Corollary 1.14.19. Principle of deformation.
Assume \(\gamma\) is a closed piecwise smooth and simple path with counterclockwise orientation, and assume \(\gamma_1, \gamma_2,\dots, \gamma_n\) are closed piecewise smooth and simple curves lying within \(\gamma\) and oriented counterclockwise. If \(f\) is holomorphic on the set \(U\) of all points lying inside or on \(\gamma\) and outside or on \(\gamma_k\) for all \(k\text{,}\) then
\begin{equation}
\int_\gamma f\, dz=\sum_{k=1}^n\int_{\gamma_k}f\, dz\text{.}\tag{1.59}
\end{equation}
We state Theorem 1.14.18 and Corollary 1.14.19 in this level of generality as a convenience, but hasten to add that in any situation that comes up in our investigations, we will be able to derive the result rigorously simply by using Theorem 1.14.8 and Theorem 1.14.12. The following example is typical.
Example 1.14.21. Principle of deformation.
Let \(\gamma(t)=3e^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\)
- Show, using only Theorem 1.14.8 and Theorem 1.14.12, that \(\int_{\gamma}1/(z-1)\, dz=\int_{\gamma'}1/(z-1)\, dz\text{,}\) where \(\gamma'(t)=1+e^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\)
- Conclude that \(\int_{\gamma}1/(z-1)\, dz=2\pi i\text{.}\)
- Express the real and imaginary parts of \(\int_{\gamma}1/(z-1)\, dz\) as real integrals over \([0,2\pi]\) to derive some interesting integral formulas.
Solution.
- Note first that \(f\) is holomorphic on \(\C-\{1\}\text{.}\) Let \(\alpha\) and \(\beta\) be the closed paths starting at the point \(3\) pictured in Figure 1.14.22. Since \(\alpha\) is closed and lies within the elementary region \(U=\C-\{1-ti\colon t\geq 0\}\text{,}\) and since \(\beta\) is closed and lies within the elementary region \(V=\C-\{1+ti\colon t\geq 0\}\text{,}\) and since \(f\) is holomorphic in both of these regions, we have\begin{equation*} \int_\alpha f\, dz=\int_\beta f\, dz=0\text{.} \end{equation*}Furthermore, letting \(\gamma''=\alpha*\beta\text{,}\) we have on the one hand\begin{equation*} \int_{\gamma''}f\, dz=\int_{\alpha} f\ , dz+\int_{\beta} f\, dz=0\text{,} \end{equation*}and on the other hand, thanks to the cancellation along the real line segments,\begin{equation*} \int_{\gamma''}f\, dz=\int_\gamma f\, dz+\int_{-\gamma'}f\, dz\text{.} \end{equation*}It follows that \(0=\int_{\gamma}f\, dz+\int_{-\gamma'}f\, dz\text{,}\) or equivalently,\begin{equation*} \int_{\gamma}f\, dz=-\int_{-\gamma'}f\, dz=\int_{\gamma'}f\, dz\text{,} \end{equation*}as claimed.
- This follows from the classic integral formula \(\int_{\gamma}1/(z-z_0)\, dz=2\pi i\text{,}\) for \(\gamma(t)=z_0+re^{it}\text{.}\)
- Using the explicit definition of the integral, we have\begin{align*} 2\pi i \amp = \int_\gamma 1/(z-1)\, dz\\ \amp = \int_0^{2\pi}\gamma'(t)/(\gamma(t)-1)\, dt\\ \amp = \int_0^{2\pi}\frac{3ie^{it}}{3e^{it}-1}\, dt\\ \amp = 3i\left( \int_0^{2\pi} \frac{3-\cos t}{10-6\cos t}\, dt-i\int_0^{2\pi}\frac{\sin t}{10-6\cos t}\, dt\right)\text{.} \end{align*}It follows that\begin{align*} \frac{2\pi}{3} \amp = \int_0^{2\pi} \frac{3-\cos t}{10-6\cos t}\, dt\\ 0 \amp = \int_0^{2\pi} \frac{\sin t}{10-6\cos t}\, dt\text{.} \end{align*}
