📂Complex Anaylsis

Proof of the Residue Theorem

Theorem ¹

Let an analytic function $f: A \subset \mathbb{C} \to \mathbb{C}$ have finitely many singularities $z_{1} , z_{2} , \cdots , z_{m}$ inside a simple closed contour $\mathscr{C}$. Then $$ \int_{\mathscr{C}} f(z) dz = 2 \pi i \sum_{k=1}^{m} \text{Res}_{z_{k}} f(z) $$

Explanation

On a first read, it is a thoroughly baffling theorem. We are supposed to compute the value of an integral, yet there is no calculus-style computation, and instead it talks about singularities and residues, so the confusion is understandable. From the statement alone, it seems to mean that we can find the value of the integral simply by computing the residues and adding them up, but can it really be so? By common sense, could there really be such an easy and convenient theorem? The answer is ‘yes’, and the residue theorem is precisely such a theorem.

Just by converting the computation of an integral into another computation, countless integrals that could not be done before become possible. Some integrals that could not be done over the reals can also be solved relatively easily by applying the residue theorem. There are truly many important theorems in complex analysis, but this one in particular yields so many useful results that it must be known.

Proof

First, let us think of $\mathscr{C}$ as split into $m$ pieces.

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

Expanding $f(z)$ as a Laurent series on each $\mathscr{C}_{k}$ gives $$ \int_{\mathscr{C}_{k}} f(z) dz = \int_{\mathscr{C}_{k}} \sum_{n = 0 }^{\infty} a_{nk} (z-z_{k}) ^{n} dz + \int_{\mathscr{C}_{k}} \sum_{n = 1 }^{\infty} { {b_{nk} } \over{ (z-z_{k}) ^{n} } } dz $$ By Cauchy’s theorem, $$ \int_{\mathscr{C}_{k}} f(z) dz = \int_{\mathscr{C}_{k}} \sum_{n = 1 }^{\infty} { {b_{nk} } \over{ (z-z_{k}) ^{n} } } dz $$ On the other hand, by Cauchy’s integral formula, $\int_{\mathscr{C}_{k}} {{1} \over {(z - z_{k})^n}} dz = \begin{cases} 2 \pi i & n = 1 \\ 0 & n \ge 2 \end{cases}$, so $$ \int_{\mathscr{C}_{k}} f(z) dz = 2 \pi i b_{1k} = 2 \pi i \text{Res}_{z_{k}} f(z) $$ Therefore, $$ \int_{\mathscr{C}} f(z) dz = 2 \pi i \sum_{k=1}^{m} \text{Res}_{z_{k}} f(z) $$

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Note

What is especially worth noting in the proof is the fact that when $n=1$, everything vanishes to $0$ except the coefficient of $\displaystyle {{1} \over {z - z_{k}}}$, that is, the residue $b_1k$. The residue theorem is so useful that if one only studies its applications, there may come a moment when one no longer even recalls why such a result comes about. If, at such a time, one can bring to mind the shapes of Cauchy’s integral formula and the Laurent expansion, then one may say one has studied with sufficient depth.