Chapter 3 • Theory & Derivations

Complex Integration, Rectifiable Curves & The Cauchy-Goursat Theorem

Line integrals along complex contours, ML-inequality, winding numbers, homotopy, and the Cauchy-Goursat theorem for simply connected domains.

§3.1 Contour Integrals along Rectifiable Arcs, The ML-Inequality & Fundamental Theorem

1. Definition of Contour Integration

Let $\gamma: [a, b] \to \mathbb{C}$ be a piecewise smooth path parameterized by $z(t) = x(t) + i y(t)$. For a continuous function $f(z) = u + iv$ on $\gamma$, the contour integral is defined by:

$$\int_\gamma f(z)\,dz = \int_a^b f(z(t)) z'(t)\,dt = \int_a^b [u x' - v y']\,dt + i \int_a^b [u y' + v x']\,dt$$

2. The ML-Inequality (Darboux Inequality)

Theorem 3.1 (The ML-Inequality): Let $\gamma$ be a rectifiable contour of arc length $L = \int_a^b |z'(t)|\,dt$. If $|f(z)| \le M$ for all $z \in \gamma$, then: $$\left| \int_\gamma f(z)\,dz \right| \le \int_\gamma |f(z)|\,|dz| \le M \cdot L$$

3. Fundamental Theorem of Complex Calculus

If $f(z)$ possesses a holomorphic primitive $F(z)$ (such that $F'(z) = f(z)$) on an open domain containing $\gamma$:

$$\int_\gamma f(z)\,dz = F(z(b)) - F(z(a))$$

In particular, if $\gamma$ is a closed loop ($z(b) = z(a)$), then $\oint_\gamma f(z)\,dz = 0$.

§3.2 Winding Numbers (Index of a Curve) & Homotopy of Paths

1. The Winding Number / Index

Let $\gamma$ be a closed piecewise smooth curve in $\mathbb{C}$, and let $z_0 \notin \gamma$. The winding number (or index) of $\gamma$ with respect to $z_0$ measures the net number of counter-clockwise revolutions that $\gamma$ makes around $z_0$:

$$\text{Ind}_\gamma(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{dz}{z - z_0}$$

Theorem 3.2: $\text{Ind}_\gamma(z_0)$ is always an integer. It is constant on each connected component of $\mathbb{C} \setminus \gamma$, and equals $0$ on the unbounded exterior component.

§3.3 The Cauchy-Goursat Theorem for Triangles, Polygons and Simply Connected Domains

1. Goursat's Lemma for Triangles

Goursat proved Cauchy's theorem without assuming continuity of partial derivatives $u_x, u_y, v_x, v_y$, requiring only complex differentiability!

Goursat's Lemma: Let $\Omega \subseteq \mathbb{C}$ be an open set and $T \subset \Omega$ a solid triangle whose boundary $\partial T$ lies in $\Omega$. If $f$ is holomorphic on $\Omega$, then: $$\oint_{\partial T} f(z)\,dz = 0$$

The proof subdivides $T$ into four congruent subtriangles, selecting the one that maximizes the integral, creating a nested sequence $T_0 \supset T_1 \supset T_2 \dots$ converging to a point $z^*$, where Taylor differentiability $f(z) = f(z^*) + f'(z^*)(z - z^*) + \epsilon(z)(z - z^*)$ forces the integral to zero.

2. The Cauchy-Goursat Theorem

Theorem 3.3 (Cauchy-Goursat Theorem): If $f(z)$ is holomorphic on a simply connected domain $D$, and $\gamma$ is any closed rectifiable Jordan contour in $D$, then: $$\oint_\gamma f(z)\,dz = 0$$

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 1 • Foundational Example 3.1: Cauchy's Integral Formula on Unit Circle

Evaluate the contour integral over the counter-clockwise circle $C: |z| = 1$:

\oint_C \frac{e^z}{z(z - 2)}\,dz

Step 1: Singularities of the integrand
The integrand $f(z) = \frac{e^z}{z(z - 2)}$ has two simple poles: at $z_1 = 0$ and $z_2 = 2$.

Step 2: Location relative to contour $C: |z| = 1$

$z_1 = 0$: lies inside $C$ since $|0| = 0 < 1$.
$z_2 = 2$: lies outside $C$ since $|2| = 2 > 1$.

Step 3: Apply Cauchy's Integral Formula
Rewrite the integrand as $\frac{g(z)}{z - 0}$ where $g(z) = \frac{e^z}{z - 2}$. Since $g(z)$ is holomorphic on and inside $C$:

\oint_C \frac{e^z}{z(z - 2)}\,dz = \oint_C \frac{g(z)}{z - 0}\,dz = 2\pi i\, g(0)

Step 4: Evaluate $g(0)$

g(0) = \frac{e^0}{0 - 2} = -\frac{1}{2}

\oint_C \frac{e^z}{z(z - 2)}\,dz = 2\pi i \left(-\frac{1}{2}\right) = -\pi i

Final Answer & Physical Insight

$-\pi i$

Tier 3 • Honors Challenge Example 3.2: Rigorous Proof of Morera's Theorem

Prove Morera's Theorem: If $f: D \to \mathbb{C}$ is continuous and $\oint_\gamma f(z)\,dz = 0$ for every closed triangular contour $\gamma \subset D$, then $f$ is holomorphic in $D$.

Step 1: Construct a local primitive $F(z)$
Let $z_0 \in D$. Choose an open disk $B(z_0, r) \subseteq D$. For any $z \in B(z_0, r)$, define:

F(z) = \int_{[z_0, z]} f(\zeta)\,d\zeta

where $[z_0, z]$ is the straight line segment from $z_0$ to $z$.

Step 2: Differentiate $F(z)$
For any $h \in \mathbb{C}$ with $z + h \in B(z_0, r)$, consider the triangle with vertices $z_0, z, z + h$. By hypothesis, the integral over the boundary of this triangle vanishes:

\int_{[z_0, z]} f(\zeta)\,d\zeta + \int_{[z, z+h]} f(\zeta)\,d\zeta + \int_{[z+h, z_0]} f(\zeta)\,d\zeta = 0

Therefore:

F(z + h) - F(z) = \int_{[z, z+h]} f(\zeta)\,d\zeta

Step 3: Difference quotient limit

\frac{F(z + h) - F(z)}{h} - f(z) = \frac{1}{h} \int_{[z, z+h]} [f(\zeta) - f(z)]\,d\zeta

\left| \frac{F(z + h) - F(z)}{h} - f(z) \right| \le \frac{1}{|h|} \cdot \epsilon \cdot |h| = \epsilon

Taking $h \to 0$ proves $F'(z) = f(z)$ for all $z \in B(z_0, r)$.

Step 4: Infinite differentiability of holomorphic functions
Since $F(z)$ is complex differentiable in $B(z_0, r)$, $F$ is holomorphic. By Corollary 4.1, every holomorphic function is infinitely differentiable! Therefore, its derivative $F'(z) = f(z)$ is also holomorphic in $B(z_0, r)$, and since $z_0$ was arbitrary, $f(z)$ is holomorphic throughout $D$. $\blacksquare$

Final Answer & Physical Insight

Proved: $F(z)$ is a primitive of $f(z) \implies F$ is holomorphic $\implies f = F'$ is holomorphic.

Tier 2 • Intermediate Exam Example 3.3: Cauchy's Integral Formula for Second Derivative

Evaluate the contour integral over counter-clockwise circle $C: |z| = 3$:

\oint_C \frac{\cos z}{(z - \pi/2)^3}\,dz

Step 1: Singularity and contour check
The integrand has a pole of order 3 at $z_0 = \pi/2 \approx 1.5708$. The contour is $|z| = 3$. Since $|\pi/2| < 3$, $z_0$ lies inside $C$.

Step 2: Apply Cauchy's formula for derivatives

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

For denominator $(z - \pi/2)^3$, we have $n + 1 = 3 \implies n = 2$. Here $f(z) = \cos z$.

Step 3: Compute second derivative

f'(z) = -\sin z, \quad f''(z) = -\cos z

Evaluate at $z_0 = \pi/2$:

f''(\pi/2) = -\cos(\pi/2) = 0

Step 4: Evaluate integral

\oint_C \frac{\cos z}{(z - \pi/2)^3}\,dz = \frac{2\pi i}{2!} f''(\pi/2) = \pi i (0) = 0

Final Answer & Physical Insight

$0$