Chapter 4 • Theory & Derivations

Cauchy's Integral Formula, Liouville's Theorem & The Maximum Modulus Principle

Cauchy's integral formulas for functions and higher derivatives, Cauchy's estimates, Liouville's theorem, Fundamental Theorem of Algebra, and the Maximum Modulus Principle.

§4.1 Cauchy's Integral Formula & Integral Formula for Higher Derivatives

1. Cauchy's Integral Formula

Theorem 4.1 (Cauchy's Integral Formula): Let $f(z)$ be holomorphic on a simply connected domain $D$, and let $\gamma$ be a simple closed counter-clockwise contour in $D$. For any point $z_0 \in \text{Int}(\gamma)$: $$f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0}\,dz$$

This remarkable theorem states that the values of a holomorphic function inside a domain are completely and uniquely determined by its values on the boundary $\gamma$!

2. Formulas for Higher Derivatives & Infinite Smoothness

Differentiating under the integral sign with respect to the parameter $z_0$:

$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}}\,dz, \quad n = 1, 2, 3, \dots$$

Corollary 4.1 (Infinite Differentiability): If $f(z)$ is holomorphic in a domain $D$, then $f$ is infinitely differentiable in $D$, and all of its derivatives $f'(z), f''(z), \dots, f^{(n)}(z)$ are also holomorphic in $D$!

§4.2 Cauchy's Estimates, Liouville's Theorem & The Fundamental Theorem of Algebra

1. Cauchy's Inequalities

Theorem 4.2 (Cauchy's Estimate): Let $f(z)$ be holomorphic on a disk $\overline{D}(z_0, R)$. If $|f(z)| \le M$ on the boundary circle $|z - z_0| = R$, then: $$|f^{(n)}(z_0)| \le \frac{n! M}{R^n}, \quad n = 0, 1, 2, \dots$$

2. Liouville's Theorem

Theorem 4.3 (Liouville's Theorem): Every bounded entire function is constant.

Proof: Let $f(z)$ be entire with $|f(z)| \le M$ for all $z \in \mathbb{C}$. For any $z_0 \in \mathbb{C}$ and any radius $R > 0$, Cauchy's estimate for $n = 1$ gives $|f'(z_0)| \le \frac{M}{R}$. Letting $R \to \infty$ yields $|f'(z_0)| = 0$. Since $z_0$ was arbitrary, $f'(z) \equiv 0$, so $f(z)$ is constant. $\blacksquare$

3. The Fundamental Theorem of Algebra

Theorem 4.4 (Fundamental Theorem of Algebra): Every non-constant polynomial $P(z) = a_n z^n + \dots + a_0$ ($a_n \ne 0, n \ge 1$) with complex coefficients has at least one root in $\mathbb{C}$.

Proof: Suppose $P(z) \ne 0$ for all $z \in \mathbb{C}$. Then $g(z) = \frac{1}{P(z)}$ is an entire function. Since $|P(z)| \to \infty$ as $|z| \to \infty$, $g(z) \to 0$, so $g(z)$ is bounded on $\mathbb{C}$. By Liouville's theorem, $g(z)$ must be constant, contradicting that $P(z)$ has degree $n \ge 1$. Thus $P(z)$ must have a root in $\mathbb{C}$. $\blacksquare$

§4.3 Morera's Theorem, The Maximum Modulus Principle & Schwarz's Lemma

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

Theorem 4.5 (Morera's Theorem): Let $f: D \to \mathbb{C}$ be continuous on a domain $D$. If $\oint_\gamma f(z)\,dz = 0$ for every closed triangular loop $\gamma \subset D$, then $f(z)$ is holomorphic in $D$.

2. The Maximum Modulus Principle

Theorem 4.6 (Maximum Modulus Principle): Let $f(z)$ be holomorphic on a bounded domain $D$ and continuous on its closure $\overline{D}$. If $|f(z)|$ attains its maximum at an interior point $z_0 \in D$, then $f(z)$ is constant throughout $D$. Consequently, the maximum of $|f(z)|$ is strictly achieved on the boundary $\partial D$: $$\max_{z \in \overline{D}} |f(z)| = \max_{z \in \partial D} |f(z)|$$

3. Schwarz's Lemma

Theorem 4.7 (Schwarz's Lemma): Let $f: \mathbb{D} \to \mathbb{D}$ be holomorphic on the unit disk $\mathbb{D} = \{|z| < 1\}$ such that $f(0) = 0$. Then: $$|f(z)| \le |z| \quad \text{for all } z \in \mathbb{D}, \quad \text{and} \quad |f'(0)| \le 1$$ If $|f(z)| = |z|$ for any non-zero $z \in \mathbb{D}$, or if $|f'(0)| = 1$, then $f(z) = e^{i\theta} z$ for some constant $\theta \in \mathbb{R}$.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 2 • Intermediate Exam Example 4.1: Proof of Liouville's Theorem via Cauchy's Estimate

Prove Liouville's theorem: Every bounded entire function is constant, using Cauchy's integral estimates.

Step 1: Hypothesis
Let $f: \mathbb{C} \to \mathbb{C}$ be an entire function (holomorphic on all of $\mathbb{C}$), and suppose there exists $M > 0$ such that $|f(z)| \le M$ for all $z \in \mathbb{C}$.

Step 2: Apply Cauchy's Estimate for $f'(z_0)$
Let $z_0 \in \mathbb{C}$ be an arbitrary point. For any $R > 0$, consider the circle $C_R = \{z \in \mathbb{C} : |z - z_0| = R\}$. By Cauchy's derivative formula:

f'(z_0) = \frac{1}{2\pi i} \oint_{C_R} \frac{f(z)}{(z - z_0)^2}\,dz

Applying the $ML$-inequality:

\left| f'(z_0) \right| \le \frac{1}{2\pi} \max_{z \in C_R} \frac{|f(z)|}{|z - z_0|^2} \cdot \text{Length}(C_R) \le \frac{1}{2\pi} \frac{M}{R^2} (2\pi R) = \frac{M}{R}

Step 3: Take the limit $R \to \infty$
Since $f$ is entire, the circle $C_R$ is valid for arbitrarily large $R$. Taking the limit as $R \to \infty$:

0 \le |f'(z_0)| \le \lim_{R \to \infty} \frac{M}{R} = 0 \implies |f'(z_0)| = 0 \implies f'(z_0) = 0

Since $z_0$ was chosen arbitrarily in $\mathbb{C}$, $f'(z) = 0$ for all $z \in \mathbb{C}$. Because $\mathbb{C}$ is connected, $f(z)$ must be constant everywhere. $\blacksquare$

Final Answer & Physical Insight

Proved: $|f'(z_0)| \le \frac{M}{R} \to 0$ as $R \to \infty \implies f'(z) \equiv 0$.

Tier 2 • Intermediate Exam Example 4.2: Singularity Classification: Isolated vs Non-Isolated Singularities

Classify all singularities of $f(z) = \frac{1}{\sin(1/z)}$ and determine whether $z = 0$ is an isolated singularity.

Step 1: Zeros of the denominator
Singularities occur where $\sin(1/z) = 0$:

\frac{1}{z} = n\pi, \quad n \in \mathbb{Z} \setminus \{0\} \implies z_n = \frac{1}{n\pi}

Step 2: Nature of singularities $z_n$
For each $n \ne 0$, $\sin(1/z_n) = 0$ and the derivative of denominator at $z_n$ is:

\left. \frac{d}{dz} \sin(1/z) \right|_{z_n} = \cos(1/z_n) \left(-\frac{1}{z_n^2}\right) = \cos(n\pi) (-n^2 \pi^2) = (-1)^{n+1} n^2 \pi^2 \ne 0

Hence, each $z_n = \frac{1}{n\pi}$ is a simple pole.

Step 3: Analysis of $z = 0$
Notice that as $n \to \infty$, the poles accumulate:

\lim_{n \to \infty} z_n = \lim_{n \to \infty} \frac{1}{n\pi} = 0

Every punctured neighborhood $D(0, \epsilon) \setminus \{0\}$ contains infinitely many poles $z_n$ (for all $n > \frac{1}{\pi\epsilon}$). Therefore, $z = 0$ is NOT an isolated singularity! It is an accumulation point of poles (non-isolated essential singularity).

Final Answer & Physical Insight

$z_n = \frac{1}{n\pi}$ ($n \in \mathbb{Z} \setminus \{0\}$) are simple poles; $z = 0$ is a non-isolated singularity (accumulation point of poles).

Tier 3 • Honors Challenge Example 4.3: Proof of the Casorati-Weierstrass Theorem

Prove the Casorati-Weierstrass Theorem: If $z_0$ is an isolated essential singularity of $f(z)$, then $f(D(z_0, \delta) \setminus \{z_0\})$ is dense in $\mathbb{C}$.

Step 1: Proof by Contradiction
Suppose the image is NOT dense in $\mathbb{C}$. Then there exists some complex number $w \in \mathbb{C}$ and $\epsilon > 0$ such that $f(z)$ never enters the $\epsilon$-neighborhood $D(w, \epsilon)$ for any $z$ in some punctured disk $D^*(z_0, \delta) = D(z_0, \delta) \setminus \{z_0\}$. That is:

|f(z) - w| \ge \epsilon \quad \text{for all } z \in D^*(z_0, \delta)

Step 2: Construct auxiliary function $g(z)$
Define:

g(z) = \frac{1}{f(z) - w}

Since $f(z) - w \ne 0$ on $D^(z_0, \delta)$, $g(z)$ is holomorphic on $D^(z_0, \delta)$. Furthermore, $g(z)$ is bounded on $D^*(z_0, \delta)$:

|g(z)| = \frac{1}{|f(z) - w|} \le \frac{1}{\epsilon}

Step 3: Apply Riemann's Removable Singularity Theorem
Since $g(z)$ is bounded on $D^*(z_0, \delta)$, the singularity at $z_0$ is removable! Thus $g(z)$ can be extended to a holomorphic function on the entire disk $D(z_0, \delta)$.

Step 4: Analyze roots of $g(z)$ at $z_0$

Case A: $g(z_0) \ne 0$. Then $\lim_{z \to z_0} f(z) = w + \frac{1}{g(z_0)}$ exists and is finite, meaning $z_0$ was a removable singularity of $f(z)$. Contradiction!
Case B: $g(z_0) = 0$ with zero of order $m \ge 1$. Then $g(z) = (z - z_0)^m h(z)$ with $h(z_0) \ne 0$, so $f(z) - w = \frac{1}{(z - z_0)^m h(z)}$ has a pole of order $m$ at $z_0$. Contradiction!

Both cases contradict that $z_0$ is an essential singularity. Hence the image must be dense in $\mathbb{C}$. $\blacksquare$

Final Answer & Physical Insight

Proved: Boundedness of $\frac{1}{f(z)-w}$ implies removable/pole, contradicting essential singularity.