The Central Idea
If \(f\) is analytic on an appropriate region containing a closed contour and its interior, then
\[ \boxed{\oint_C f(z)\,dz=0.} \]Cauchy's integral formula goes further. If \(z_0\) lies inside a positively oriented simple closed contour \(C\), then
\[ \boxed{ f(z_0)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z-z_0}\,dz. } \]The Cauchy–Goursat Theorem
A standard version states that if \(f\) is analytic throughout a simply connected domain \(D\), then every closed piecewise smooth contour \(C\subset D\) satisfies
\[ \boxed{\oint_C f(z)\,dz=0.} \]Goursat's refinement is important because continuity of \(f'\) is not required.
Simply Connected Domains
Informally, a simply connected domain has no holes. Every closed curve can be continuously contracted to a point without leaving the domain.
Examples include open disks, half-planes, rectangles, and \(\mathbb{C}\).
Why Holes Matter
The punctured plane
\[ \mathbb{C}\setminus\{0\} \]is not simply connected. A circle surrounding the origin cannot be shrunk to a point without crossing the missing point.
This explains why
\[ \oint_{|z|=1}\frac1z\,dz=2\pi i \]does not contradict Cauchy's theorem.
Analyticity Must Cover the Relevant Interior
The basic theorem cannot be applied if a singularity lies inside the contour.
For example, \(1/z\) is analytic on the unit circle but not at the enclosed point \(z=0\). That interior singularity is the obstruction.
Example: Entire Integrand
For any closed contour \(C\), evaluate
\[ \oint_C(z^4+e^z+\cos z)\,dz. \]Each term is entire, so the integrand is analytic everywhere. Therefore,
\[ \boxed{0}. \]Path Independence
If \(f\) is analytic on a simply connected domain, then two contours with the same endpoints give the same integral.
\[ \boxed{ \int_{C_1}f(z)\,dz=\int_{C_2}f(z)\,dz. } \]Traversing \(C_1\) and then \(C_2\) in reverse creates a closed contour, so Cauchy's theorem forces the difference of the two integrals to vanish.
Analytic Functions and Antiderivatives
On a simply connected domain, every analytic function has an antiderivative.
Fix a point \(z_0\) and define
\[ F(z)=\int_{z_0}^{z}f(\zeta)\,d\zeta. \]Path independence makes this definition well-defined, and
\[ \boxed{F'(z)=f(z).} \]Contour Deformation
If \(C_1\) can be continuously deformed into \(C_2\) through a region where \(f\) remains analytic, then under the usual hypotheses
\[ \boxed{ \oint_{C_1}f(z)\,dz=\oint_{C_2}f(z)\,dz. } \]Singularities are what prevent such deformations.
Cauchy's Integral Formula
Let \(C\) be a positively oriented simple closed contour, let \(f\) be analytic on an open set containing \(C\) and its interior, and let \(z_0\) lie inside \(C\). Then
\[ \boxed{ f(z_0)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z-z_0}\,dz. } \]Equivalently,
\[ \boxed{ \oint_C\frac{f(z)}{z-z_0}\,dz=2\pi i\,f(z_0). } \]Why the Integral Formula Is Remarkable
The value of an analytic function at an interior point is determined by its values along the surrounding boundary.
Boundary information controls interior information. This principle is one of the deepest recurring themes in complex analysis.
Recognizing the Pattern
Look for an integrand of the form
\[ \frac{f(z)}{z-z_0}. \]If \(f\) is analytic on and inside the contour and \(z_0\) lies inside, the integral is
\[ \boxed{2\pi i\,f(z_0).} \]Example: Exponential Numerator
Let \(C\) be the positively oriented circle \(|z|=2\). Evaluate
\[ \oint_C\frac{e^z}{z-1}\,dz. \]The point \(1\) lies inside \(C\), and \(e^z\) is entire. Therefore,
\[ \boxed{2\pi i e.} \]Example: Polynomial Numerator
Let \(C\) be \(|z|=3\). Evaluate
\[ \oint_C\frac{z^2+4z+1}{z-i}\,dz. \]Since \(|i|=1<3\), \(i\) lies inside. Let
\[ f(z)=z^2+4z+1. \]Then
\[ f(i)=i^2+4i+1=4i. \]Hence
\[ \oint_C\frac{f(z)}{z-i}\,dz = 2\pi i(4i) = \boxed{-8\pi}. \]What If \(z_0\) Lies Outside?
If \(z_0\) lies outside \(C\), then
\[ \frac{f(z)}{z-z_0} \]is analytic on and inside \(C\), assuming \(f\) is.
Therefore, by Cauchy's theorem,
\[ \boxed{ \oint_C\frac{f(z)}{z-z_0}\,dz=0. } \]Example: Exterior Singular Point
Evaluate
\[ \oint_{|z|=1}\frac{e^z}{z-2}\,dz. \]The singular point \(2\) lies outside the unit circle, so the integrand is analytic on and inside the contour.
\[ \boxed{0}. \]Orientation
Cauchy's integral formula is stated for positive, counterclockwise orientation. Reversing orientation changes the sign:
\[ \boxed{ \oint_{-C}\frac{f(z)}{z-z_0}\,dz=-2\pi i\,f(z_0). } \]Cauchy's Formula for Derivatives
The integral formula extends to every derivative:
\[ \boxed{ f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz. } \]Equivalently,
\[ \boxed{ \oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz = \frac{2\pi i}{n!}f^{(n)}(z_0). } \]The First-Derivative Case
\[ \boxed{ f'(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z-z_0)^2}\,dz. } \]The derivative at an interior point is determined entirely by boundary values.
Example: A Higher-Order Denominator
Evaluate
\[ \oint_{|z|=2}\frac{e^z}{z^3}\,dz. \]Here \((z-0)^3=(z-0)^{2+1}\), so \(n=2\). Since
\[ (e^z)''=e^z \]and \(e^0=1\),
\[ \oint_{|z|=2}\frac{e^z}{z^3}\,dz = \frac{2\pi i}{2!} = \boxed{\pi i}. \]Analytic Functions Are Infinitely Differentiable
Because every derivative can be represented by a contour integral, a holomorphic function possesses derivatives of every order.
\[ \boxed{ \text{holomorphic} \Rightarrow \text{infinitely complex differentiable}. } \]Connection to Taylor Series
Cauchy's derivative formulas lead naturally to
\[ \boxed{ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n } \]within a disk limited by the nearest obstruction to analyticity.
Cauchy Estimates
Suppose \(f\) is analytic on and inside
\[ |z-z_0|=R \]and
\[ |f(z)|\le M \]on the circle. Then
\[ \boxed{ |f^{(n)}(z_0)| \le \frac{n!M}{R^n}. } \]Example: Bound a Derivative
Suppose \(f\) is analytic on and inside \(|z|=3\) and \(|f(z)|\le10\) there.
Then
\[ |f''(0)| \le \frac{2!\cdot10}{3^2} = \boxed{\frac{20}{9}}. \]The Mean-Value Property
On a circle centered at \(z_0\), Cauchy's formula can be rewritten as
\[ \boxed{ f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0+Re^{it})\,dt. } \]The value at the center equals the average of the function around the boundary circle.
A Preview of Liouville's Theorem
Cauchy estimates imply that a bounded entire function must be constant:
\[ \boxed{ f\text{ entire and bounded} \Rightarrow f\text{ constant}. } \]This result gives an elegant route to the Fundamental Theorem of Algebra.
A Preview of the Maximum Modulus Principle
A nonconstant analytic function cannot attain a genuine interior maximum of its modulus on a connected open region.
This is another way boundary behavior constrains the interior.
A Converse Direction: Morera's Theorem
Cauchy's theorem says analyticity produces zero integrals around suitable closed contours.
Morera's theorem gives a converse under continuity: if a continuous function has zero integral around every triangle in a domain, then it is analytic there.
Integrands with Several Linear Factors
Consider
\[ \frac{1}{(z-a)(z-b)}. \]If only one of \(a\) and \(b\) lies inside the contour, absorb the other factor into the analytic numerator and apply Cauchy's integral formula.
Example: One Enclosed Singular Point
Let \(C\) be \(|z|=2\). Evaluate
\[ \oint_C\frac{1}{(z-1)(z-4)}\,dz. \]Only \(z=1\) lies inside. Set
\[ f(z)=\frac1{z-4}. \]Then \(f\) is analytic on and inside \(C\), and
\[ f(1)=-\frac13. \]Therefore,
\[ \boxed{ -\frac{2\pi i}{3}. } \]Matching Higher Powers
If the denominator is
\[ (z-z_0)^m, \]then the derivative order is
\[ n=m-1. \]Hence
\[ \boxed{ \oint_C\frac{f(z)}{(z-z_0)^m}\,dz = \frac{2\pi i}{(m-1)!}f^{(m-1)}(z_0). } \]Example: Derivative Formula with a Shift
Evaluate
\[ \oint_C\frac{z^3}{(z-1)^2}\,dz \]where \(C\) positively encloses \(z=1\).
Here \(f(z)=z^3\) and \(n=1\). Since
\[ f'(1)=3, \]we get
\[ \boxed{6\pi i}. \]Common Errors
- Ignoring interior singularities: analyticity on the contour alone is not enough.
- Forgetting domain topology: a hole can obstruct the simplest form of Cauchy's theorem.
- Using the integral formula for an exterior point: if the singular point is outside and the integrand is otherwise analytic, use Cauchy's theorem instead.
- Forgetting orientation: clockwise traversal reverses the sign.
- Forgetting \(n!\): higher-derivative formulas include a factorial.
- Confusing theorem and formula: Cauchy's theorem gives a vanishing integral; the integral formula evaluates a specific \(1/(z-z_0)\)-type integrand.
A Reliable Procedure: Cauchy's Theorem
- Identify the contour and its interior.
- Locate every singular point of the integrand.
- Check whether any singularity lies on or inside the contour.
- Verify the domain conditions for the theorem being used.
- If the hypotheses hold, conclude that the integral is zero.
A Reliable Procedure: Cauchy's Integral Formula
- Identify the factor \(z-z_0\).
- Check that \(z_0\) lies inside the contour.
- Verify that the numerator \(f(z)\) is analytic on and inside.
- Confirm orientation.
- Evaluate \(f(z_0)\).
- Multiply by \(2\pi i\).
A Reliable Procedure: Cauchy's Derivative Formula
- Match the denominator to \((z-z_0)^{n+1}\).
- Determine \(n\).
- Identify the analytic numerator \(f(z)\).
- Compute \(f^{(n)}(z_0)\).
- Multiply by \(2\pi i/n!\).
Worked Example: Cauchy's Integral Formula
Evaluate
\[ \oint_{|z|=4}\frac{z^2+1}{z-2}\,dz. \]The point \(2\) lies inside the circle. Let \(f(z)=z^2+1\). Then
\[ f(2)=5. \]Therefore,
\[ \boxed{10\pi i}. \]Worked Example: Higher Derivative
Evaluate
\[ \oint_{|z|=2}\frac{\sin z}{z^4}\,dz. \]Since \(z^4=(z-0)^{3+1}\), use \(n=3\).
For \(f(z)=\sin z\),
\[ f^{(3)}(z)=-\cos z, \qquad f^{(3)}(0)=-1. \]Hence
\[ \boxed{ -\frac{\pi i}{3}. } \]Practice Problems
Practice 1
Evaluate
\[ \oint_C(3z^2+e^z)\,dz \]for any closed contour \(C\).
Practice 2
Evaluate
\[ \oint_{|z|=3}\frac{z^2}{z-1}\,dz. \]Practice 3
Evaluate
\[ \oint_{|z|=2}\frac{e^z}{z-i}\,dz. \]Practice 4
Evaluate
\[ \oint_{|z|=1}\frac1{z-3}\,dz. \]Practice 5
Evaluate
\[ \oint_{|z|=2}\frac{e^z}{z^2}\,dz. \]Practice 6
Evaluate
\[ \oint_{|z|=2}\frac{\cos z}{z^3}\,dz. \]Practice 7
Let \(C\) be \(|z-2|=3\). Evaluate
\[ \oint_C\frac{z+1}{z-i}\,dz. \]Practice 8
Evaluate
\[ \oint_C\frac1z\,dz \]when \(C\) is the circle \(|z|=2\) traversed clockwise.
Practice 9
Suppose \(f\) is analytic on and inside \(|z|=4\) and \(|f(z)|\le6\) on the circle. Bound \(|f^{(3)}(0)|\).
Practice 10
Which theorem directly evaluates
\[ \oint_C\frac{f(z)}{z-z_0}\,dz \]when \(f\) is analytic on and inside \(C\) and \(z_0\) lies inside?
Practice Answers
Practice 1
\[ \boxed{0}. \]Practice 2
\[ \boxed{2\pi i}. \]Practice 3
\[ \boxed{2\pi i e^i}. \]Practice 4
\[ \boxed{0}. \]Practice 5
\[ \boxed{2\pi i}. \]Practice 6
Since \(f''(0)=-1\),
\[ \boxed{-\pi i}. \]Practice 7
Since \(|i-2|=\sqrt5<3\), the point \(i\) lies inside. Thus
\[ \boxed{2\pi i(1+i)}. \]Practice 8
\[ \boxed{-2\pi i}. \]Practice 9
\[ |f^{(3)}(0)| \le \frac{3!\cdot6}{4^3} = \boxed{\frac{9}{16}}. \]Practice 10
\[ \boxed{\text{Cauchy's integral formula}.} \]Checking Strategies
- Locate singularities before choosing a theorem.
- Check whether each singularity is inside, outside, or on the contour.
- Verify positive or negative orientation.
- For the integral formula, isolate \(z-z_0\).
- For higher powers, match the exponent to \(n+1\).
- Remember the factorial in the derivative formula.
- If an answer is unexpectedly nonzero, inspect the interior for an obstruction to analyticity.
Connections to Earlier Mathematics
- Complex integration: analyticity controls closed-contour integrals.
- Topology: holes and singularities affect deformation and path independence.
- Differentiation: derivatives of every order arise from contour integrals.
- Power series: Cauchy's formulas lead to Taylor expansions.
- Estimation: Cauchy estimates control derivative size from boundary bounds.
- Algebra: Liouville's theorem leads to the Fundamental Theorem of Algebra.
What to Remember
Under suitable hypotheses,
\[ \boxed{\oint_Cf(z)\,dz=0.} \]If \(z_0\) lies inside a positively oriented contour,
\[ \boxed{ f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-z_0}\,dz. } \]For higher derivatives,
\[ \boxed{ f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz. } \]And on \(|z-z_0|=R\), if \(|f(z)|\le M\), then
\[ \boxed{ |f^{(n)}(z_0)| \le \frac{n!M}{R^n}. } \]The central structural idea is: for analytic functions, boundary information controls the interior with extraordinary precision.
Where We Go Next
Cauchy's formulas explain why analytic functions possess derivatives of every order and why local power-series expansions are unavoidable.
The next lesson, Series and Singularities, develops Taylor and Laurent series, classifies removable singularities, poles, and essential singularities, and prepares the machinery for Residues.