📂Complex Anaylsis

Zeros and Poles of Meromorphic Functions

Theorem ¹

Let $f$ be a function that is analytic on a simple closed curve $\mathscr{C}$, having $Z$ zeros and $P$ poles inside $\mathscr{C}$, and suppose $f(z) \ne 0$ on $\mathscr{C}$. 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 numbers counted with all multiplicities.

Explanation

Analytic Number Theory

If a function $f : \mathbb{C} \to \mathbb{C}$ has no poles, this becomes a formula for finding the number of solutions of the equation $f(z) = 0$. The point worth noting is that an integer has appeared. At first glance, complex analysis seems to have nothing whatsoever to do with number theory, but in reality it is used very frequently. Indeed, in number theory there is even a term, ’elementary proof’, which refers to a proof that does not use complex analysis.

The Log Trick

You might wonder how an integrand shaped like that could ever be useful, but it is the form of the derivative of $\log f$. This kind of shape appears more often than one might think throughout mathematics, regardless of the field.

Derivative of an arithmetic function: The derivative or derived function $f '$ 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, suppose that $f$ has just one zero of order $n$ and one pole of order $m$. Geometrically these are single points, but algebraically they carry multiplicities $n$ and $m$ respectively; that is, several points are clustered together in one place. We derive the formula only for these two points, and for multiple points we generalize by splitting up $\mathscr{C}$.

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Part 1. Zeros

If $\alpha$ is a zero of $f$ of order $n$, then for some function $g$ it can be expressed as $f(z) = (z- \alpha)^{n} g(z)$, and dividing its derivative $$ f ' (z) = n \left( n - \alpha \right)^{n-1} g(z) + (z- \alpha)^{n} g ' (z) $$ by $f(z)$, the left-hand side can be represented in the form of the derivative-of-a-logarithmic-composite-function trick as follows. $$ {{f ' (z)} \over {f(z)}} = {{n} \over {z - \alpha}} + {{g ' (z)} \over {g(z)}} $$ Since $f(z) \ne 0$ on $\mathscr{C}$, $\log f$ is analytic over the whole of $\mathscr{C}$. On the other hand, since the derivative of an analytic function is analytic, $f ' / f$ is also analytic, and likewise, since $f(z) \ne 0$ on $\mathscr{C}$, $1 / (z - \alpha)$ is also analytic. Accordingly, $\displaystyle {{g ' (z)} \over {g(z)}} = {{n} \over {z - \alpha}} - {{f ' (z)} \over {f(z)}}$ is also analytic at $\alpha$.

• Cauchy 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 contains no other zeros or poles of $f$, by the Cauchy integral formula we have $$ \int_{\mathcal{N}_{\alpha}} {{n} \over {z - \alpha}} dz = 2 n \pi i $$ and by Cauchy’s theorem we have $$ \int_{\mathcal{N}_{\alpha}} {{g ' (z)} \over {g(z)}} dz = 0 $$ so as a result we obtain the following. $$ \int_{\mathcal{N}_{\alpha}} {{f ' (z)} \over {f(z)}} dz = 0 + 2 n \pi i $$

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Part 2. Poles

For poles it is similar to what we did for zeros. Let $\beta$ be a pole of $f$ of order $m$. Then for some function $h$ it can be expressed as $\displaystyle f(z) = {{h(z)} \over {(z - \beta)^m}}$.

On the other hand, in $\displaystyle {{f ' (z)} \over {f(z)}} = {{h ' (z)} \over {h(z)}} - {{m} \over {z - \beta}}$, since $\displaystyle {{h ' (z)} \over {h(z)}}$ is analytic in the neighborhood $\mathcal{N}_{\beta}$ of $\beta$, by Cauchy’s theorem $$ \int_{\mathcal{N}_{\beta}} {{h ' (z)} \over {h(z)}} dz = 0 $$ and by the Cauchy 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 deformation lemma, repeating the calculations up to this point finitely many times for all zeros and poles, we obtain 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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