📂Complex Anaylsis

Zeros and Poles of Meromorphic Functions

Theorem ¹

In a simple closed path $\mathscr{C}$, let an analytic function $f$ have $Z$ zeros and $P$ poles inside $\mathscr{C}$, and on $\mathscr{C}$, let it be $f(z) \ne 0$. Then, $${{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f ' (z)} \over {f(z)}} dz = Z - P$$

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• $Z$ and $P$ are the sums including multiplicities.

Explanation

Analytic Number Theory

If the function $f : \mathbb{C} \to \mathbb{C}$ has no poles, it would become a formula for counting the number of solutions to the equation $f(z) = 0$. It’s noteworthy that integers have appeared. At first glance, complex analysis might seem completely unrelated to number theory, yet it’s extensively used in reality. In number theory, there’s even a term ’elementary proof’ referring to proofs that don’t utilize complex analysis.

Log Trick

It might seem pointless to use a function shaped like this, but it’s the derivative form of $\log f$. Such a form appears more often in mathematics than one might think, across various fields.

Derivative of an Arithmetic Function: The derivative or derivative form of an arithmetic function $f$ is defined as follows. $$f ' (n) := f(n) \log n \qquad , n \in \mathbb{N}$$

Proof

Strategy: Without loss of generality, assume that there is only one $n$th-order zero and one $m$th-order pole. Geometrically, they might seem like a single point, but algebraically, they are considered to have multiplicities $n$ and $m$, i.e., multiple points clustered together. Derive a formula for these two points and generalize it by partitioning $\mathscr{C}$ for multiple points.

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Part 1. Zeros

Let $\alpha$ be a $f$’s $n$th-order zero. Then, for some function $g$, it can be expressed as $f(z) = (z- \alpha)^{n} g(z)$, and its derivative $$f ' (z) = n \left( n - \alpha \right)^{n-1} g(z) + (z- \alpha)^{n} g ' (z)$$ when divided by $f(z)$, can be represented as a logarithmic derivative trick form. $${{f ' (z)} \over {f(z)}} = {{n} \over {z - \alpha}} + {{g ' (z)} \over {g(z)}}$$ Since $f(z) \ne 0$ on $\mathscr{C}$, $\log f$ is analytic over the whole of $\mathscr{C}$. Furthermore, since an analytic function’s derivative is also analytic, $f ' / f$ is also analytic, and similarly, since $f(z) \ne 0$ on $\mathscr{C}$, $1 / (z - \alpha)$ is also analytic. Therefore, $\displaystyle {{g ' (z)} \over {g(z)}} = {{n} \over {z - \alpha}} - {{f ' (z)} \over {f(z)}}$ is also analytic at $\alpha$.

• Cauchy’s Integral Formula $$f(\alpha) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz$$
• Cauchy’s Theorem: $$\int_{\mathscr{C}} f(z) dz = 0$$

In some neighborhood $\mathcal{N}_{\alpha}$ of $\alpha$ that does not include any other zeros or poles of $f$, according to Cauchy’s Integral Formula, $$\int_{\mathcal{N}_{\alpha}} {{n} \over {z - \alpha}} dz = 2 n \pi i$$ and by Cauchy’s Theorem, $$\int_{\mathcal{N}_{\alpha}} {{g ' (z)} \over {g(z)}} dz = 0$$ thus obtaining the following result. $$\int_{\mathcal{N}_{\alpha}} {{f ' (z)} \over {f(z)}} dz = 0 + 2 n \pi i$$

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Part 2. Poles

The case for poles is similar to that for zeros. Let $\beta$ be a $f$’s $m$th-order pole. Then, for some function $h$, it can be expressed as $\displaystyle f(z) = {{h(z)} \over {(z - \beta)^m}}$.

Meanwhile, near $\beta$, $\displaystyle {{f ' (z)} \over {f(z)}} = {{h ' (z)} \over {h(z)}} - {{m} \over {z - \beta}}$ is analytic, so by Cauchy’s Theorem, $$\int_{\mathcal{N}_{\beta}} {{h ' (z)} \over {h(z)}} dz = 0$$ and by Cauchy’s Integral Formula, $$\int_{\mathcal{N}_{\beta}} {{m} \over {z - \alpha}} dz = 2 m \pi i$$

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Part 3. Conclusion

By the Generalized Contraction Lemma, repeating the calculations finitely for all zeros and poles yields the following. $${{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f ' (z)} \over {f(z)}} dz = \sum n_{i} - \sum m_{j} = Z - P$$

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