Theorem 4.4.1.
If \(f\) is holomorphic in an open set containing \(\overline D[w,R]\) then
\begin{equation*}
f(w) \ = \ \frac 1 {2 \pi i} \int_{C[w,R]} \frac{ f(z) }{
z-w } \, \diff{z} \, \text{.}
\end{equation*}
Section 4.4 Cauchy’s Integral Formula
We recall our notations
\begin{align*}
C[a,r] \amp = \left\{ z \in \C : \abs{ z-a } = r
\right\}\\
D[a,r] \amp = \left\{z \in \C: \abs{z-a}\lt r
\right\}\\
\overline D[a,r] \amp = \left\{z \in \C: \abs{z-a} \le r
\right\}
\end{align*}
for the circle, open disk, and closed disk, respectively, with center \(a \in \C\) and radius \(r > 0\text{.}\) Unless stated otherwise, we orient \(C[a,r]\) counter-clockwise.
This is Cauchy’s Integral Formula for the case that the integration path is a circle; we will prove the general statement at the end of this chapter. However, already this special case is worth meditating over: the data on the right-hand side of Theorem 4.4.1 is entirely given by the values that \(f(z)\) takes on for \(z\) on the circle \(C[w,R]\text{.}\) Thus Cauchy’s Integral Formula says that this data determines \(f(w)\text{.}\) This has the flavor of mean-value theorems, which the following corollary makes even more apparent.
Corollary 4.4.2.
If \(f=u+iv\) is holomorphic in an open set containing \(\overline D[w,R]\text{,}\) then
\begin{align*}
f(w) \amp \ = \ \frac 1 {2 \pi} \int_0^{2 \pi} f \left( w + R \,
e^{it} \right) \diff{t} \, ,\\
u(w) \amp \ = \ \frac{1}{2 \pi} \int_0^{2 \pi} u \left( w + R \,
e^{it}\right) \diff{t} \, ,\\
v(w) \amp \ = \
\frac 1 {2 \pi} \int_0^{2 \pi} v \left( w + R \, e^{it}
\right) \diff{t} \,\text{.}
\end{align*}
Proof.
By assumption, \(f\) is holomorphic in an open set \(G\) that contains \(\overline D[w,R]\text{,}\) and so \(\frac{ f(z) }{ z-w }\) is holomorphic in \(H := G
\setminus \{ w \}\text{.}\) For any \(0 \lt r \lt R\text{,}\)
\begin{equation*}
C[w,r] \sim_H C[w,R] \,\text{,}
\end{equation*}
and so Cauchy’s Theorem 4.3.4 and Exercise 4.5.4 give
\begin{align}
\amp \left| \int_{C[w,R]} \frac{ f(z) }{ z-w } \, \diff{z} -
2 \pi i \, f(w) \right| \notag\\
\amp \qquad = \ \left| \int_{C[w,r]} \frac{
f(z) }{ z-w } \, \diff{z} - f(w) \int_{C[w,r]} \frac{ \diff{z}
}{ z-w } \right| \nonumber\notag\\
\amp \qquad = \ \left| \int_{C[w,r]} \frac{ f(z) - f(w) }{ z-w
} \, \diff{z} \right| \nonumber\notag\\
\amp \qquad \le \ \max_{
z \in C[w,r] } \left| \frac{ f(z) - f(w) }{ z-w } \right|
\length \left( C[w,r] \right)\tag{4.6}\\
\amp \qquad = \ \max_{ z \in C[w,r] } \frac{ \left| f(z) -
f(w) \right| }{ r } \ 2 \pi r \nonumber\notag\\
\amp \qquad = \ 2 \pi \max_{ z \in C[w,r] } \left| f(z) - f(w)
\right| \, . \nonumber\notag
\end{align}
Now let \(\epsilon > 0\text{.}\) Because \(f\) is continuous at \(w\text{,}\) there exists \(\delta > 0\) such that \(|z-w| \lt
\delta\) implies
\begin{equation*}
\left| f(z) - f(w) \right| \ \lt \ \frac \epsilon {2 \pi} \,\text{.}
\end{equation*}
In particular, this will hold for \(z \in C[w, \frac \delta 2]\text{,}\) and so (4.6) implies, with \(r = \frac \delta 2\text{,}\)
\begin{equation*}
\left| \int_{C[w,R]} \frac{ f(z) }{ z-w } \, \diff{z} - 2 \pi
i \, f(w) \right| \ \lt \ \epsilon \, \text{.}
\end{equation*}
Since we can choose \(\epsilon\) as small as we’d like, the left-hand side must be zero, which proves Theorem 4.4.1.
Corollary 4.4.2 now follows by definition of the complex integral:
\begin{equation*}
f(w) \ = \ \frac 1 {2 \pi i} \int_0^{2 \pi} \frac{ f \left( w
+ R \, e^{it} \right) }{ w + R \, e^{it} - w } \, i R \,
e^{it} \, \diff{t} \ = \ \frac 1 {2 \pi} \int_0^{2 \pi} f
\left( w + R \, e^{it} \right) \diff{t} \, \text{,}
\end{equation*}
which splits into real and imaginary parts as
\begin{equation*}
u(w) + i \, v(w) \ = \ \frac{1}{2 \pi} \int_0^{2 \pi} u \left(
w + R \, e^{it}\right) \diff{t} \ + \ i \, \frac 1 {2 \pi}
\int_0^{2 \pi} v \left( w + R \, e^{it} \right) \diff{t} \,
.
\end{equation*}
Theorem 4.4.1 can be used to compute integrals of a certain nature.
Example 4.4.3.
We’d like to determine
\begin{equation*}
\int_{ C[i,1] } \frac{ \diff{z} }{ z^2 + 1 } \,\text{.}
\end{equation*}
The function \(f(z) = \frac{ 1 }{ z+i }\) is holomorphic in \(\C \setminus \{ -i \}\text{,}\) which contains \(\overline D[i,1]\text{.}\) Thus we can apply Theorem 4.4.1:
\begin{equation*}
\int_{ C[i,1] } \frac{ \diff{z} }{ z^2 + 1 } \ = \ \int_{
C[i,1] } \frac{ \frac{ 1 }{ z+i } }{ z-i } \, \diff{z} \ = \ 2
\pi i \, f(i) \ = \ 2 \pi i \, \frac{ 1 }{ 2i } \ = \ \pi \, \text{.}
\end{equation*}
Now we would like to extend Theorem 4.4.1 by replacing \(C[w,R]\) with any simple closed piecewise smooth path \(\gg\) around \(w\text{.}\) Intuitively, Cauchy’s Theorem 4.3.4 should supply such an extension: assuming that \(f\) is holomorphic in a region \(G\) that includes \(\gg\) and its inside, we can find a small \(R\) such that \(\overline D[w,R]
\subseteq G\text{,}\) and since \(\frac{ f(z) }{ z-w }\) is holomorphic in \(H := G \setminus \{ w \}\) and \(\gg \sim_H C[w,R]\text{,}\) Theorem 4.3.4 and Theorem 4.4.1 yield
\begin{equation*}
f(w) \ = \ \frac 1 {2 \pi i} \int_\gg \frac{ f(z) }{ z-w } \,
\diff{z} \, \text{.}
\end{equation*}
This all smells like good coffee, except ... we might be just dreaming. The argument may be intuitively clear, but intuition doesn’t prove anything. We’ll look at it carefully, fill in the gaps, and then we’ll see what we have proved.
First, we need a notion of the inside of a simple closed path. The fact that any such path \(\gg\) divides the complex plane into two connected open sets of \(\gg\) (the bounded one of which we call the inside or interior of \(\gg\)) is one of the first substantial theorems ever proved in topology, the Jordan Curve Theorem, due to Camille Jordan (1838–1922). In this book we shall assume the validity of the Jordan Curve Theorem.
1
This is the Jordan of Jordan normal form fame, but not the one of Gauß–Jordan elimination.
Second, we need to specify the orientation of \(\gg\text{,}\) since if the formula gives \(f(w)\) for one orientation then it will give \(-f(w)\) for the other orientation.
Definition 4.4.4.
A piecewise smooth simple closed path \(\gg\) is positively oriented if it is parametrized so that its inside is on the left as our parametrization traverses \(\gamma\text{.}\) An example is a counter-clockwise oriented circle.
Third, if \(\gg\) is positively oriented and \(\overline D[w,R]\) is a closed disk inside \(\gg\) then we need a homotopy from \(\gg\) to the counterclockwise circle \(C[w,R]\) that stays inside \(\gg\) and away from \(D[w,R]\text{.}\) This is provided directly by another substantial theorem of topology, the Annulus Theorem, although there are other methods. Again, in this book we shall assume the existence of this homotopy.
These results of topology seem intuitively obvious but are surprisingly difficult to prove. If you’d like to see a proof, we recommend that you take a course in topology.
There is yet another subtle problem. We assumed that \(\gg\) is in \(G\text{,}\) but we also need the interior of \(\gg\) to be contained in \(G\text{,}\) since we need to apply Cauchy’s Theorem to the homotopy between \(\gg\) and \(C[w,R]\text{.}\) We could just add this as an assumption to our theorem, but the following formulation will be more convenient later.
Theorem 4.4.5. Cauchy’s Integral Formula.
Suppose \(f\) is holomorphic in the region \(G\) and \(\gg\) is a positively oriented, simple, closed, piecewise smooth path, such that \(w\) is inside \(\gg\) and \(\gg \sim_G 0\text{.}\) Then
\begin{equation*}
f(w) \ = \ \frac 1 {2 \pi i} \int_\gg \frac{ f(z) }{ z-w } \,
\diff{z} \, \text{.}
\end{equation*}
So all that we need to finish the proof of Theorem 4.4.5 is one more fact from topology. But we can prove this one:
Proposition 4.4.6.
Suppose \(\gg\) is a simple, closed, piecewise smooth path in the region \(G\text{.}\) Then \(G\) contains the interior of \(\gg\) if and only if \(\gg\sim_G0\text{.}\)
Proof.
One direction is easy: If \(G\) contains the interior of \(\gg\) and \(\overline D[w,R]\) is any closed disk in the interior of \(\gg\) then there is a \(G\)-homotopy from \(\gg\) to \(C[w,R]\text{,}\) and \(C[w,R] \sim_G 0\text{.}\)
In the other direction, we argue by contradiction: Assume \(\gg\sim_G0\) but \(G\) does not contain the interior of \(\gg\text{.}\) So we can find a point \(w\) in the interior of \(\gg\) which is not in \(G\text{.}\)
Define \(g(z)=\frac1{z-w}\) for \(z\ne w\text{.}\) Now \(g\) is holomorphic on \(G\) and \(\gg\sim_G0\text{,}\) so Corollary 4.3.8 applies, and we have \(\int_\gg g(z)\,\diff{z} = 0\text{.}\) On the other hand, choose \(R>0\) so that \(\overline D[w,R]\) is inside \(\gg\text{.}\) There is a homotopy in \(\C\setminus \{\,w\,\}\) from \(\gg\) to \(C[w,R]\text{,}\) so Cauchy’s Theorem 4.3.4, plus Exercise 4.5.4, shows that \(\int_\gg g(z)\,\diff{z} = 2\pi i\text{.}\)
This contradiction finishes the proof.
Notice that, instead of using topology to prove a theorem about holomorphic functions, we just used holomorphic functions to prove a theorem about topology.
Example 4.4.7.
Continuing Example 4.4.3, Theorem 4.4.5 says that
\begin{equation*}
\int_\gg \frac{ \diff{z} }{ z^2 + 1 } \ = \ \pi
\end{equation*}
for any positively oriented, simple, closed, piecewise smooth path \(\gg\) that contains \(i\) on its inside and that is \(( \C \setminus \{ -i \})\)-contractible.
Example 4.4.8.
To compute
\begin{equation*}
\int_{ C[0,3] } \frac{ \exp(z) }{ z^2 - 2z } \, \diff{z}
\end{equation*}
we use the partial fractions expansion from Example 4.3.11:
\begin{equation*}
\int_{ C[0,3] } \frac{ \exp(z) }{ z^2 - 2z } \, \diff{z} \ = \
\frac 1 2 \int_{ C[0,3] } \frac{ \exp(z) }{ z - 2 } \,
\diff{z} \ - \ \frac 1 2 \int_{ C[0,3] } \frac{ \exp(z) }{ z
} \, \diff{z} \, \text{.}
\end{equation*}
For the two integrals on the right-hand side, we can use Theorem 4.4.1 with the function \(f(z) = \exp(z)\text{,}\) which is entire, and so (note that both 2 and 0 are inside \(\gg\))
\begin{equation*}
\int_{ C[0,3] } \frac{ \exp(z) }{ z^2 - 2z } \, \diff{z} \ = \
\frac 1 2 \, 2 \pi i \cdot \exp(2) - \frac 1 2 \, 2 \pi i
\cdot \exp(0) \ = \ \pi i \left( e^2 - 1 \right) \,\text{.}
\end{equation*}
