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Course 5

Cauchy Integral Formula & Derivatives

Master Cauchy's integral formula, learn derivative formulas for analytic functions, understand Morera's theorem, and explore the mean value property.

Intermediate5-6 hours10 Practice Questions

Learning Objectives

Understand and apply Cauchy's integral formula

Derive formulas for derivatives of analytic functions

Master Morera's theorem and its applications

Understand the mean value property

Learn higher derivative formulas

Apply these formulas to evaluate integrals

Understand the relationship between analyticity and integral representation

Work with derivative estimates

Cauchy's Integral Formula

Theorem 5.1: Cauchy's Integral Formula

Let

f(z)

be analytic in a domain

D

, and let

C

be a simple closed curve in

D

with interior also in

D

. Then for any point

z_0

inside

C

f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-z_0}dz

Proof of Theorem 5.1:

Consider a small circle $C_\varepsilon: |z-z_0| = \varepsilon$ inside $C$ . Since $\frac{f(z)}{z-z_0}$ is analytic in the region between $C$ and $C_\varepsilon$ , by deformation:

\oint_C \frac{f(z)}{z-z_0}dz = \oint_{C_\varepsilon} \frac{f(z)}{z-z_0}dz

Parameterizing $C_\varepsilon$ as $z = z_0 + \varepsilon e^{i\theta}$ :

\oint_{C_\varepsilon} \frac{f(z)}{z-z_0}dz = \int_0^{2\pi} \frac{f(z_0 + \varepsilon e^{i\theta})}{\varepsilon e^{i\theta}} i\varepsilon e^{i\theta}d\theta = i\int_0^{2\pi} f(z_0 + \varepsilon e^{i\theta})d\theta

As $\varepsilon \to 0$ , by continuity: $= 2\pi i f(z_0)$ , giving the result.

∎

Corollary 5.1: Mean Value Property

f

is analytic and

C: |z-z_0| = R

is a circle in the domain, then:

f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + Re^{i\theta})d\theta

Example 5.1

Problem: Evaluate $\oint_{|z|=2} \frac{e^z}{z-1}dz$ .

Solution:

Since $e^z$ is analytic everywhere and $z=1$ is inside $|z|=2$ , by Cauchy's integral formula:

\oint_{|z|=2} \frac{e^z}{z-1}dz = 2\pi i \cdot e^1 = 2\pi i e

Derivative Formulas

Theorem 5.2: Derivative Formula

Under the same conditions as Theorem 5.1,

f

has derivatives of all orders at

z_0

, and:

f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-z_0)^{n+1}}dz

for

n = 1, 2, 3, \ldots

Proof of Theorem 5.2:

Differentiate Cauchy's integral formula with respect to $z_0$ :

f'(z_0) = \frac{d}{dz_0}\left[\frac{1}{2\pi i} \oint_C \frac{f(z)}{z-z_0}dz\right] = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z-z_0)^2}dz

Continuing this process gives the formula for higher derivatives. The differentiation under the integral sign is justified by uniform convergence.

∎

Example 5.2

Problem: Find $f'(i)$ if $f(z) = \frac{1}{2\pi i} \oint_{|z|=2} \frac{e^z}{z-z_0}dz$ for $|z_0| < 2$ .

Solution:

By Cauchy's formula, $f(z_0) = e^{z_0}$ for $|z_0| < 2$ , so $f'(i) = e^i$ .

Alternatively, using the derivative formula: $f'(i) = \frac{1}{2\pi i} \oint_{|z|=2} \frac{e^z}{(z-i)^2}dz = e^i$ .

Theorem 5.3: Morera's Theorem

f

is continuous in a domain

D

and

\oint_C f(z)dz = 0

for every closed curve

C

D

, then

f

is analytic in

D

Remark 5.1

Morera's theorem is the converse of Cauchy's theorem. It provides a useful criterion for proving analyticity when direct verification of the Cauchy-Riemann equations is difficult.

Morera's Theorem (Converse of Cauchy's Theorem)

Theorem 5.4: Morera's Theorem

f

is continuous in a simply connected domain

D

and

\oint_C f(z)dz = 0

for every closed curve

C

D

, then

f

is analytic in

D

Remark 5.1

Morera's theorem provides a necessary and sufficient condition for analyticity in simply connected domains. Together with Cauchy's theorem, it shows that analyticity is equivalent to the vanishing of contour integrals.

Liouville's Theorem and Applications

Theorem 5.5: Liouville's Theorem

f

is entire (analytic in the entire complex plane) and bounded, then

f

is constant.

Proof of Theorem 5.5:

Suppose $|f(z)| \leq M$ for all $z$ . For any $z_0$ , use the derivative formula with a large circle:

|f'(z_0)| = \left|\frac{1}{2\pi i} \oint_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^2}dz\right| \leq \frac{1}{2\pi} \cdot \frac{M}{R^2} \cdot 2\pi R = \frac{M}{R}

Since this holds for arbitrarily large $R$ , we have $f'(z_0) = 0$ for all $z_0$ , so $f$ is constant.

∎

Example 5.3

Problem: Show that a non-constant entire function cannot be bounded.

Solution:

This is a direct consequence of Liouville's theorem. If an entire function is bounded, it must be constant. Therefore, any non-constant entire function (like $e^z$ , $\sin z$ ) is unbounded.

Corollary 5.3: Fundamental Theorem of Algebra

Every non-constant polynomial

P(z)

has at least one complex root.

Proof of Corollary 5.3:

Suppose $P(z) \neq 0$ for all $z$ . Then $1/P(z)$ is entire. Since $|P(z)| \to \infty$ as $|z| \to \infty$ , we have $|1/P(z)| \to 0$ , so $1/P(z)$ is bounded. By Liouville's theorem, $1/P(z)$ is constant, hence $P(z)$ is constant, a contradiction.

∎

Example 5.4

Problem: Use Cauchy's integral formula to show that if $f$ is analytic in $|z| \leq R$ and $|f(z)| \leq M$ , then $|f^{(n)}(0)| \leq \frac{n!M}{R^n}$ .

Solution:

Using the derivative formula on $|z| = R$ :

$|f^{(n)}(0)| = \left|\frac{n!}{2\pi i} \oint_{|z|=R} \frac{f(z)}{z^{n+1}}dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}$

Significance of Cauchy's Integral Formula

Representation of Analytic Functions

The formula shows that the value of an analytic function at any point is determined by its values on surrounding curves, demonstrating the "rigidity" of analytic functions.

Infinite Differentiability

The formula for derivatives shows that analytic functions are infinitely differentiable, a remarkable property not shared by real differentiable functions.

Connection to Power Series

The derivative formulas are essential for proving that analytic functions have power series representations, connecting complex analysis to series theory.

Maximum Modulus Principle

Theorem 5.6: Maximum Modulus Principle

f

is analytic and non-constant in a domain

D

, then

|f(z)|

cannot attain a maximum at any interior point of

D

. If

D

is bounded and

f

is continuous on

\overline{D}

, then the maximum occurs on the boundary

\partial D

Proof of Theorem 5.6:

Suppose $|f|$ attains a maximum at an interior point $z_0$ . By the mean value property:

|f(z_0)| = \left|\frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})d\theta\right| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})|d\theta \leq |f(z_0)|

For equality to hold, $|f|$ must be constant on every circle centered at $z_0$ , which implies $f$ is constant (by analytic continuation).

∎

Example 5.5

Problem: Find the maximum of $|z^2 + 1|$ on the closed disk $|z| \leq 2$ .

Solution:

Since $z^2 + 1$ is entire, by the maximum modulus principle, the maximum occurs on the boundary $|z| = 2$ .

On the boundary, $z = 2e^{i\theta}$ , so $|z^2 + 1| = |4e^{i2\theta} + 1|$

Maximum occurs when $4e^{i2\theta} = 4$ (aligned with 1), giving maximum $|4 + 1| = 5$

Theorem 5.7: Minimum Modulus Principle

f

is analytic and non-constant in a domain

D

, and

f(z) \neq 0

for all

z \in D

, then

|f(z)|

cannot attain a minimum at any interior point of

D

Corollary 5.4: Open Mapping Theorem

f

is analytic and non-constant in a domain

D

, then

f(D)

is an open set. In other words, analytic functions map open sets to open sets.

Additional Worked Examples

Example 5.6

Problem: Use Cauchy's integral formula to show that if $f$ is entire and bounded, then $f$ is constant.

Solution:

This is Liouville's theorem. For any $z_0$ , use the derivative formula with a large circle of radius $R$ :

$|f'(z_0)| = \left|\frac{1}{2\pi i} \oint_{|z-z_0|=R} \frac{f(z)}{(z-z_0)^2}dz\right| \leq \frac{1}{2\pi} \cdot \frac{M}{R^2} \cdot 2\pi R = \frac{M}{R}$

Since this holds for arbitrarily large $R$ , $f'(z_0) = 0$ , so $f$ is constant.

Example 5.7

Problem: Evaluate $\oint_{|z|=2} \frac{\sin z}{z^3}dz$ .

Solution:

The function $\frac{\sin z}{z^3}$ has a pole at $z = 0$ . Using the derivative formula:

$\oint_{|z|=2} \frac{\sin z}{z^3}dz = 2\pi i \cdot \frac{\sin''(0)}{2!} = 2\pi i \cdot \frac{-\sin(0)}{2} = 0$

Actually, since $\sin z = z - \frac{z^3}{6} + \cdots$ , we have $\frac{\sin z}{z^3} = \frac{1}{z^2} - \frac{1}{6} + \cdots$ , so the residue (coefficient of $1/z$ ) is 0.

Example 5.8

Problem: Prove that if $f$ is analytic in $|z| < R$ and $|f(z)| \leq M$ , then $|f^{(n)}(0)| \leq \frac{n!M}{R^n}$ .

Solution:

Using the derivative formula on a circle of radius $r < R$ :

$|f^{(n)}(0)| = \left|\frac{n!}{2\pi i} \oint_{|z|=r} \frac{f(z)}{z^{n+1}}dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{r^{n+1}} \cdot 2\pi r = \frac{n!M}{r^n}$

Taking $r \to R$ gives the result.

Study Tips

1. Understanding Cauchy's Formula

The formula expresses the value of an analytic function at a point as an integral over surrounding curves. This is remarkable: the entire function is determined by its boundary values.

2. Computing Derivatives

Use the derivative formula when the integrand has the form $\frac{f(z)}{(z-z_0)^{n+1}}$ where $f$ is analytic. The $n$ th derivative is given by $\frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_0)^{n+1}}dz$ .

3. Liouville's Theorem Applications

If you can show a function is entire and bounded, it must be constant. This is often used to prove identities or show that certain functions don't exist.

Additional Worked Examples

Example 5.9

Problem: Use Cauchy's integral formula to evaluate $\oint_{|z|=2} \frac{e^z}{z-1}dz$ .

Solution:

The function $e^z$ is analytic inside $|z| = 2$ , and we have the form $\frac{f(z)}{z-1}$ where $f(z) = e^z$ .

By Cauchy's integral formula: $\oint_{|z|=2} \frac{e^z}{z-1}dz = 2\pi i \cdot e^1 = 2\pi i e$

Example 5.10

Problem: Find $f''(0)$ if $f(z) = \frac{1}{2\pi i} \oint_{|z|=1} \frac{e^{\zeta}}{\zeta^3}d\zeta$ .

Solution:

Using the derivative formula: $f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta$

Here $n = 2$ , $z_0 = 0$ , and $f(\zeta) = e^{\zeta}$

$f''(0) = \frac{2!}{2\pi i} \oint_{|z|=1} \frac{e^{\zeta}}{\zeta^3}d\zeta = \frac{2}{2\pi i} \cdot 2\pi i \cdot \frac{e^0}{2!} = 1$

Example 5.11

Problem: Prove that if $f$ is analytic and $|f(z)| \leq |z|^n$ for large $|z|$ , then $f$ is a polynomial of degree at most $n$ .

Solution:

Using the derivative estimate with $f^{(n+1)}(0)$ and a large circle, we find $|f^{(n+1)}(0)| = 0$ .

By uniqueness of power series, all coefficients beyond degree $n$ are zero, so $f$ is a polynomial.

Example 5.12

Problem: Show that if $f$ is entire and $\text{Re}(f(z)) \leq 0$ for all $z$ , then $f$ is constant.

Solution:

Consider $g(z) = e^{f(z)}$ . Since $\text{Re}(f(z)) \leq 0$ , we have $|g(z)| = e^{\text{Re}(f(z))} \leq 1$ .

$g$ is entire and bounded, so by Liouville's theorem, $g$ is constant.

Therefore $f$ is constant.

Example 5.13

Problem: Use Cauchy's integral formula to show that if $f$ is analytic and $|f(z)| \leq M$ for $|z| \leq R$ , then $|f^{(n)}(0)| \leq \frac{n!M}{R^n}$ .

Solution:

Using the derivative formula on $|z| = R$ :

$|f^{(n)}(0)| = \left|\frac{n!}{2\pi i} \oint_{|z|=R} \frac{f(z)}{z^{n+1}}dz\right| \leq \frac{n!}{2\pi} \cdot \frac{M}{R^{n+1}} \cdot 2\pi R = \frac{n!M}{R^n}$

Example 5.14

Problem: Evaluate $\oint_{|z|=2} \frac{e^z}{(z-1)^2}dz$ .

Solution:

Using the derivative formula with $n = 1$ :

$\oint_{|z|=2} \frac{e^z}{(z-1)^2}dz = 2\pi i \cdot f'(1)$ where $f(z) = e^z$

Since $f'(z) = e^z$ , we have $f'(1) = e$

Therefore: $\oint_{|z|=2} \frac{e^z}{(z-1)^2}dz = 2\pi i e$

Example 5.15

Problem: Prove that if $f$ is entire and $f(z+1) = f(z)$ for all $z$ , and $|f(z)|$ is bounded in the strip $0 \leq \text{Re } z \leq 1$ , then $f$ is constant.

Solution:

By periodicity, $f$ is bounded everywhere. By Liouville's theorem, $f$ is constant.

Example 5.16

Problem: Show that if $f$ is analytic in $|z| < R$ and $|f(z)| \leq M$ , then $|f'(0)| \leq M/R$ .

Solution:

Using the derivative formula with $n = 1$ and a circle of radius $r < R$ :

$|f'(0)| = \left|\frac{1}{2\pi i} \oint_{|z|=r} \frac{f(z)}{z^2}dz\right| \leq \frac{1}{2\pi} \cdot \frac{M}{r^2} \cdot 2\pi r = \frac{M}{r}$

Taking $r \to R$ gives the result.

Example 5.17

Problem: Show that if $f$ is analytic in $|z| \leq R$ and $|f(z)| \leq M$ , then $|f(z)| \leq M$ for $|z| < R$ with equality only if $f$ is constant.

Solution:

This follows from the maximum modulus principle. If $|f|$ attains its maximum at an interior point, then $f$ must be constant.

Example 5.18

Problem: Evaluate $\oint_{|z|=1} \frac{\cos z}{z^2}dz$ .

Solution:

Using the derivative formula with $n = 1$ :

$\oint_{|z|=1} \frac{\cos z}{z^2}dz = 2\pi i \cdot \cos'(0) = 2\pi i \cdot (-\sin(0)) = 0$

Example 5.19

Problem: Use Cauchy's integral formula to evaluate $\oint_{|z|=1} \frac{e^{2z}}{z}dz$ .

Solution:

The function $e^{2z}$ is entire. By Cauchy's integral formula:

$\oint_{|z|=1} \frac{e^{2z}}{z}dz = 2\pi i \cdot e^{2 \cdot 0} = 2\pi i$

Example 5.20

Problem: Show that an entire function $f$ with $|f(z)| \leq |z|^n$ for large $|z|$ is a polynomial of degree at most $n$ .

Solution:

Using the derivative estimate on large circles, we find that $f^{(n+1)}(0) = 0$ . Since $f$ is entire, its Taylor series is valid everywhere, and all coefficients beyond degree $n$ are zero.

Example 5.21

Problem: Show that if $f$ is analytic and $|f(z)| = 1$ on $|z| = 1$ , and $f(0) = 0$ , then $f(z) = cz$ for some constant $|c| = 1$ .

Solution:

Consider $g(z) = f(z)/z$ for $z \neq 0$ . Then $|g(z)| = 1$ on $|z| = 1$ .

By maximum modulus principle, $|g(z)| \leq 1$ for $|z| \leq 1$ , and since $|g| = 1$ on the boundary, $g$ is constant.

Therefore $f(z) = cz$ with $|c| = 1$ .

Example 5.22

Problem: Prove that if $f$ is entire and $|f(z)| \leq 1 + |z|^{1/2}$ , then $f$ is constant.

Solution:

Using the derivative estimate with $n = 1$ on large circles, we find that $f'(0) = 0$ .

Since $f$ is entire, shifting the argument shows $f'(z) = 0$ for all $z$ , so $f$ is constant.

Example 5.23

Problem: Evaluate $\oint_{|z|=1} \frac{e^{z^2}}{z^3}dz$ .

Solution:

Using the derivative formula with $n = 2$ and $f(z) = e^{z^2}$ :

$\oint_{|z|=1} \frac{e^{z^2}}{z^3}dz = 2\pi i \cdot \frac{f''(0)}{2!} = \pi i \cdot 2e^0 = 2\pi i$

Example 5.24

Problem: Show that if $f$ is entire and $\text{Re}(f(z)) \geq 0$ for all $z$ , then $f$ is constant.

Solution:

Consider $g(z) = e^{-f(z)}$ . Then $|g(z)| = e^{-\text{Re}(f(z))} \leq 1$ .

Since $g$ is entire and bounded, by Liouville's theorem, $g$ is constant, so $f$ is constant.

Example 5.25

Problem: Prove that a non-constant entire function cannot be bounded away from zero. That is, if $f$ is entire and $|f(z)| \geq \varepsilon > 0$ for all $z$ , then $f$ is constant.

Solution:

If $|f(z)| \geq \varepsilon > 0$ for all $z$ , then $1/f(z)$ is entire and $|1/f(z)| \leq 1/\varepsilon$ .

By Liouville's theorem, since $1/f$ is entire and bounded, it must be constant.

Therefore, $f$ is constant. This shows that non-constant entire functions must take values arbitrarily close to zero.

Example 5.26

Problem: Use the maximum modulus principle to show that if $f$ is analytic in a bounded domain $D$ and continuous on $\overline{D}$ , then the maximum of $|f(z)|$ occurs on the boundary $\partial D$ , unless $f$ is constant.

Solution:

Suppose the maximum occurs at an interior point $z_0 \in D$ . By the maximum modulus principle, if $|f|$ attains its maximum at an interior point, then $f$ must be constant in a neighborhood of $z_0$ .

Since $D$ is connected, the identity theorem (which we'll see later) implies that $f$ is constant throughout $D$ .

Therefore, if $f$ is not constant, the maximum must occur on the boundary $\partial D$ .

This is a fundamental result in complex analysis with applications to harmonic functions, potential theory, and optimization problems.

Practice Problems

Cauchy Integral Formula & Derivatives Practice

Questions

Correct

Accuracy

Cauchy's integral formula states that if

f(z)

is analytic inside and on a simple closed curve

C

, then for

z_0

inside

C

Easy

Not attempted

Using Cauchy's integral formula,

\oint_{|z|=2} \frac{e^z}{z-1}dz

equals:

Medium

Not attempted

The

n

-th derivative formula for analytic functions is:

Medium

Not attempted

Morera's theorem states that if

\oint_C f(z)dz = 0

for every closed curve

C

in a domain

D

, then:

Medium

Not attempted

The mean value property states that for an analytic function

f

and circle

C: |z-z_0| = R

Hard

Not attempted

Using the derivative formula,

\oint_{|z|=1} \frac{\cos z}{z^3}dz

equals:

Hard

Not attempted

An important consequence of Cauchy's integral formula is that:

Easy

Not attempted

f

is analytic and

|f(z)| \leq M

|z-z_0| = R

, then

|f^{(n)}(z_0)| \leq

Hard

Not attempted

The formula

f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + Re^{i\theta})d\theta

is called the:

Medium

Not attempted

To evaluate

\oint_{|z|=2} \frac{z^2}{z-1}dz

using Cauchy's formula, we get:

Medium

Not attempted

Frequently Asked Questions

Why is Cauchy's integral formula so powerful?

It allows us to compute values of analytic functions and their derivatives at interior points using only information on the boundary. This is remarkable because in general, boundary values don't determine interior values, but for analytic functions they do.

How do I use Cauchy's integral formula to evaluate integrals?

If you have an integral of the form $\oint \frac{f(z)}{z-z_0}dz$ where $f$ is analytic, then the integral equals $2\pi i f(z_0)$ if $z_0$ is inside the curve, and $0$ if it's outside.

Why do analytic functions have derivatives of all orders?

This follows from the derivative formulas. Since we can repeatedly differentiate under the integral sign in Cauchy's formula, all derivatives exist. This is much stronger than real differentiability, where a function can be differentiable but not twice differentiable.

What is Liouville's theorem and why is it important?

Liouville's theorem states that every bounded entire function is constant. This is a powerful result with many applications: it proves the Fundamental Theorem of Algebra, shows that non-constant entire functions are unbounded, and is used to prove many other theorems in complex analysis.

What does the mean value property tell us?

It shows that analytic functions behave like harmonic functions: the value at a point equals the average of values on any circle centered at that point. This leads to important properties like the maximum modulus principle.