📂Complex Anaylsis

Contraction Lemma for Complex Path Integrals

Theorem¹

Let $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic at every point except a point $\alpha$ in the interior of $\mathscr{C}$, on a simply connected region containing the simple closed path $\mathscr{C}$. Then for a closed curve $\mathscr{C} '$ centered at $\alpha$ in the interior of $\mathscr{C}$, $$ \int_{\mathscr{C}} f(z) dz = \int_{\mathscr{C} '} f(z) dz $$

Explanation

The statement is long, but in the end it says that when performing a complex integral over a closed path, we can contract that closed path around some point.

Being able to change the interval of integration this freely is unimaginable for real numbers. Note that $f$ does not necessarily have to be non-differentiable at $\alpha$. Also, as you will see from the proof, there is no reason why $\mathscr{C} '$ must be a circle.

Proof

$$\displaystyle \int_{\Gamma_{1} } f(z) dz + \int_{\Gamma_{2} } f(z) dz = \int_{\mathscr{C}} f(z) dz - \int_{\mathscr{C} '} f(z) dz$$

Cauchy–Goursat theorem: If $f$ is analytic on a simply connected region $\mathscr{R}$, then for a simple closed path ${\Gamma}$ inside $\mathscr{R}$, $$ \int_{{\Gamma}} f(z) dz = 0 $$

By the Cauchy–Goursat theorem, $\displaystyle \int_{\Gamma_{1} } f(z) dz = 0$ and $\displaystyle \int_{\Gamma_{2} } f(z) dz =0$. Therefore $$ \int_{\mathscr{C}} f(z) dz = \int_{\mathscr{C} '} f(z) dz $$

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Generalization

Contraction lemma generalized for partitions: Let $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic at every point except finitely many points $\alpha_{1} , \alpha_{2}, \cdots \alpha_{n}$ in the interior of $\mathscr{C}$, on a simply connected region containing the simple closed path $\mathscr{C}$. Then for circles $\mathscr{C_k}$ centered at $\alpha_{k}$ in the interior of $\mathscr{C}$, $$ \int_{\mathscr{C}} f(z) dz = \sum_{k=1}^{n} \int_{\mathscr{C}_{k}} f(z) dz$$

By applying the idea of splitting paths a bit further, we naturally obtain the generalized theorem.