📂Complex Anaylsis

Contraction Lemma for Complex Path Integrals

Theorem ¹

Let’s say in a simply connected domain containing a simple closed path $\mathscr{C}$, $f: A \subseteq \mathbb{C} \to \mathbb{C}$ is analytic at all points excluding point $\alpha$ inside $\mathscr{C}$. Then, for a closed curve $\mathscr{C} '$ centered at $\alpha$ inside $\mathscr{C}$, $$ \int_{\mathscr{C}} f(z) dz = \int_{\mathscr{C} '} f(z) dz $$

Explanation

It’s a long way to say, but essentially, it means that when doing complex integration over a closed path, you can contract that closed path around some point.

It’s unthinkable in real numbers to freely change the integration interval like this. Note that it’s not necessary for $\alpha$ to be non-differentiable. Moreover, as we can see in the proof process, there’s no reason why $\mathscr{C} '$ must be a circle necessarily.

Proof

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

According to the Cauchy-Goursat theorem, in a simply connected domain $\mathscr{R}$ if $f$ is analytic, then for a simple closed path ${\Gamma}$ inside $\mathscr{R}$, $$ \int_{{\Gamma}} f(z) dz = 0 $$

From the Cauchy-Goursat theorem, it follows that $\displaystyle \int_{\Gamma_{1} } f(z) dz = 0$ 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

Generalized contraction subsidiary lemma for division: In a simply connected domain that includes a simple closed path $\mathscr{C}$, suppose $f: A \subseteq \mathbb{C} \to \mathbb{C}$ is analytic at all points excluding a finite number of points $\alpha_{1} , \alpha_{2}, \cdots \alpha_{n}$ inside $\mathscr{C}$. Then, for a circular path $\mathscr{C_k}$ centered at $\alpha_{k}$ inside $\mathscr{C}$, $$ \int_{\mathscr{C}} f(z) dz = \sum_{k=1}^{n} \int_{\mathscr{C}_{k}} f(z) dz$$

By applying the idea of dividing the path a bit further, a naturally generalized theorem can be obtained.