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We will shortly turn to the complex-analytic approach to multiplicative number theory, which relies on the basic properties of complex analytic functions. In this supplement to the main notes, we quickly review the portions of complex analysis that we will be using in this course. We will not attempt a comprehensive review of this subject; for instance, we will completely neglect the conformal geometry or Riemann surface aspect of complex analysis, and we will also avoid using the various boundary convergence theorems for Taylor series or Dirichlet series (the latter type of result is traditionally utilised in multiplicative number theory, but I personally find them a little unintuitive to use, and will instead rely on a slightly different set of complex-analytic tools). We will also focus on the “local” structure of complex analytic functions, in particular adopting the philosophy that such functions behave locally like complex polynomials; the classical “global” theory of entire functions, while traditionally used in the theory of the Riemann zeta function, will be downplayed in these notes. On the other hand, we will play up the relationship between complex analysis and Fourier analysis, as we will incline to using the latter tool over the former in some of the subsequent material. (In the traditional approach to the subject, the Mellin transform is used in place of the Fourier transform, but we will not emphasise the role of the Mellin transform here.)

We begin by recalling the notion of a holomorphic function, which will later be shown to be essentially synonymous with that of a complex analytic function.

Definition 1 (Holomorphic function) Let ${\Omega}$ be an open subset of ${{\bf C}}$, and let ${f: \Omega \rightarrow {\bf C}}$ be a function. If ${z \in {\bf C}}$, we say that ${f}$ is complex differentiable at ${z}$ if the limit

$\displaystyle f'(z) := \lim_{h \rightarrow 0; h \in {\bf C} \backslash \{0\}} \frac{f(z+h)-f(z)}{h}$

exists, in which case we refer to ${f'(z)}$ as the (complex) derivative of ${f}$ at ${z}$. If ${f}$ is differentiable at every point ${z}$ of ${\Omega}$, and the derivative ${f': \Omega \rightarrow {\bf C}}$ is continuous, we say that ${f}$ is holomorphic on ${\Omega}$.

Exercise 2 Show that a function ${f: \Omega \rightarrow {\bf C}}$ is holomorphic if and only if the two-variable function ${(x,y) \mapsto f(x+iy)}$ is continuously differentiable on ${\{ (x,y) \in {\bf R}^2: x+iy \in \Omega\}}$ and obeys the Cauchy-Riemann equation

$\displaystyle \frac{\partial}{\partial x} f(x+iy) = \frac{1}{i} \frac{\partial}{\partial y} f(x+iy). \ \ \ \ \ (1)$

Basic examples of holomorphic functions include complex polynomials

$\displaystyle P(z) = a_n z^n + \dots + a_1 z + a_0$

as well as the complex exponential function

$\displaystyle \exp(z) := \sum_{n=0}^\infty \frac{z^n}{n!}$

which are holomorphic on the entire complex plane ${{\bf C}}$ (i.e., they are entire functions). The sum or product of two holomorphic functions is again holomorphic; the quotient of two holomorphic functions is holomorphic so long as the denominator is non-zero. Finally, the composition of two holomorphic functions is holomorphic wherever the composition is defined.

Exercise 3

• (i) Establish Euler’s formula

$\displaystyle \exp(x+iy) = e^x (\cos y + i \sin y)$

for all ${x,y \in {\bf R}}$. (Hint: it is a bit tricky to do this starting from the trigonometric definitions of sine and cosine; I recommend either using the Taylor series formulations of these functions instead, or alternatively relying on the ordinary differential equations obeyed by sine and cosine.)

• (ii) Show that every non-zero complex number ${z}$ has a complex logarithm ${\log(z)}$ such that ${\exp(\log(z))=z}$, and that this logarithm is unique up to integer multiples of ${2\pi i}$.
• (iii) Show that there exists a unique principal branch ${\hbox{Log}(z)}$ of the complex logarithm in the region ${{\bf C} \backslash (-\infty,0]}$, defined by requiring ${\hbox{Log}(z)}$ to be a logarithm of ${z}$ with imaginary part between ${-\pi}$ and ${\pi}$. Show that this principal branch is holomorphic with derivative ${1/z}$.

In real analysis, we have the fundamental theorem of calculus, which asserts that

$\displaystyle \int_a^b F'(t)\ dt = F(b) - F(a)$

whenever ${[a,b]}$ is a real interval and ${F: [a,b] \rightarrow {\bf R}}$ is a continuously differentiable function. The complex analogue of this fact is that

$\displaystyle \int_\gamma F'(z)\ dz = F(\gamma(1)) - F(\gamma(0)) \ \ \ \ \ (2)$

whenever ${F: \Omega \rightarrow {\bf C}}$ is a holomorphic function, and ${\gamma: [0,1] \rightarrow \Omega}$ is a contour in ${\Omega}$, by which we mean a piecewise continuously differentiable function, and the contour integral ${\int_\gamma f(z)\ dz}$ for a continuous function ${f}$ is defined via change of variables as

$\displaystyle \int_\gamma f(z)\ dz := \int_0^1 f(\gamma(t)) \gamma'(t)\ dt.$

The complex fundamental theorem of calculus (2) follows easily from the real fundamental theorem and the chain rule.

In real analysis, we have the rather trivial fact that the integral of a continuous function on a closed contour is always zero:

$\displaystyle \int_a^b f(t)\ dt + \int_b^a f(t)\ dt = 0.$

In complex analysis, the analogous fact is significantly more powerful, and is known as Cauchy’s theorem:

Theorem 4 (Cauchy’s theorem) Let ${f: \Omega \rightarrow {\bf C}}$ be a holomorphic function in a simply connected open set ${\Omega}$, and let ${\gamma: [0,1] \rightarrow \Omega}$ be a closed contour in ${\Omega}$ (thus ${\gamma(1)=\gamma(0)}$). Then ${\int_\gamma f(z)\ dz = 0}$.

Exercise 5 Use Stokes’ theorem to give a proof of Cauchy’s theorem.

A useful reformulation of Cauchy’s theorem is that of contour shifting: if ${f: \Omega \rightarrow {\bf C}}$ is a holomorphic function on a open set ${\Omega}$, and ${\gamma, \tilde \gamma}$ are two contours in an open set ${\Omega}$ with ${\gamma(0)=\tilde \gamma(0)}$ and ${\gamma(1) = \tilde \gamma(1)}$, such that ${\gamma}$ can be continuously deformed into ${\tilde \gamma}$, then ${\int_\gamma f(z)\ dz = \int_{\tilde \gamma} f(z)\ dz}$. A basic application of contour shifting is the Cauchy integral formula:

Theorem 6 (Cauchy integral formula) Let ${f: \Omega \rightarrow {\bf C}}$ be a holomorphic function in a simply connected open set ${\Omega}$, and let ${\gamma: [0,1] \rightarrow \Omega}$ be a closed contour which is simple (thus ${\gamma}$ does not traverse any point more than once, with the exception of the endpoint ${\gamma(0)=\gamma(1)}$ that is traversed twice), and which encloses a bounded region ${U}$ in the anticlockwise direction. Then for any ${z_0 \in U}$, one has

$\displaystyle \int_\gamma \frac{f(z)}{z-z_0}\ dz= 2\pi i f(z_0).$

Proof: Let ${\varepsilon > 0}$ be a sufficiently small quantity. By contour shifting, one can replace the contour ${\gamma}$ by the sum (concatenation) of three contours: a contour ${\rho}$ from ${\gamma(0)}$ to ${z_0+\varepsilon}$, a contour ${C_\varepsilon}$ traversing the circle ${\{z: |z-z_0|=\varepsilon\}}$ once anticlockwise, and the reversal ${-\rho}$ of the contour ${\rho}$ that goes from ${z_0+\varepsilon}$ to ${\gamma_0}$. The contributions of the contours ${\rho, -\rho}$ cancel each other, thus

$\displaystyle \int_\gamma \frac{f(z)}{z-z_0}\ dz = \int_{C_\varepsilon} \frac{f(z)}{z-z_0}\ dz.$

By a change of variables, the right-hand side can be expanded as

$\displaystyle 2\pi i \int_0^1 f(z_0 + \varepsilon e^{2\pi i t})\ dt.$

Sending ${\varepsilon \rightarrow 0}$, we obtain the claim. $\Box$

The Cauchy integral formula has many consequences. Specialising to the case when ${\gamma}$ traverses a circle ${\{ z: |z-z_0|=r\}}$ around ${z_0}$, we conclude the mean value property

$\displaystyle f(z_0) = \int_0^1 f(z_0 + re^{2\pi i t})\ dt \ \ \ \ \ (3)$

whenever ${f}$ is holomorphic in a neighbourhood of the disk ${\{ z: |z-z_0| \leq r \}}$. In a similar spirit, we have the maximum principle for holomorphic functions:

Lemma 7 (Maximum principle) Let ${\Omega}$ be a simply connected open set, and let ${\gamma}$ be a simple closed contour in ${\Omega}$ enclosing a bounded region ${U}$ anti-clockwise. Let ${f: \Omega \rightarrow {\bf C}}$ be a holomorphic function. If we have the bound ${|f(z)| \leq M}$ for all ${z}$ on the contour ${\gamma}$, then we also have the bound ${|f(z_0)| \leq M}$ for all ${z_0 \in U}$.

Proof: We use an argument of Landau. Fix ${z_0 \in U}$. From the Cauchy integral formula and the triangle inequality we have the bound

$\displaystyle |f(z_0)| \leq C_{z_0,\gamma} M$

for some constant ${C_{z_0,\gamma} > 0}$ depending on ${z_0}$ and ${\gamma}$. This ostensibly looks like a weaker bound than what we want, but we can miraculously make the constant ${C_{z_0,\gamma}}$ disappear by the “tensor power trick“. Namely, observe that if ${f}$ is a holomorphic function bounded in magnitude by ${M}$ on ${\gamma}$, and ${n}$ is a natural number, then ${f^n}$ is a holomorphic function bounded in magnitude by ${M^n}$ on ${\gamma}$. Applying the preceding argument with ${f, M}$ replaced by ${f^n, M^n}$ we conclude that

$\displaystyle |f(z_0)|^n \leq C_{z_0,\gamma} M^n$

and hence

$\displaystyle |f(z_0)| \leq C_{z_0,\gamma}^{1/n} M.$

Sending ${n \rightarrow \infty}$, we obtain the claim. $\Box$

Another basic application of the integral formula is

Corollary 8 Every holomorphic function ${f: \Omega \rightarrow {\bf C}}$ is complex analytic, thus it has a convergent Taylor series around every point ${z_0}$ in the domain. In particular, holomorphic functions are smooth, and the derivative of a holomorphic function is again holomorphic.

Conversely, it is easy to see that complex analytic functions are holomorphic. Thus, the terms “complex analytic” and “holomorphic” are synonymous, at least when working on open domains. (On a non-open set ${\Omega}$, saying that ${f}$ is analytic on ${\Omega}$ is equivalent to asserting that ${f}$ extends to a holomorphic function of an open neighbourhood of ${\Omega}$.) This is in marked contrast to real analysis, in which a function can be continuously differentiable, or even smooth, without being real analytic.

Proof: By translation, we may suppose that ${z_0=0}$. Let ${C_r}$ be a a contour traversing the circle ${\{ z: |z|=r\}}$ that is contained in the domain ${\Omega}$, then by the Cauchy integral formula one has

$\displaystyle f(z) = \frac{1}{2\pi i} \int_{C_r} \frac{f(w)}{w-z}\ dw$

for all ${z}$ in the disk ${\{ z: |z| < r \}}$. As ${f}$ is continuously differentiable (and hence continuous) on ${C_r}$, it is bounded. From the geometric series formula

$\displaystyle \frac{1}{w-z} = \frac{1}{w} + \frac{1}{w^2} z + \frac{1}{w^3} z^2 + \dots$

and dominated convergence, we conclude that

$\displaystyle f(z) = \sum_{n=0}^\infty (\frac{1}{2\pi i} \int_{C_r} \frac{f(w)}{w^{n+1}}\ dw) z^n$

with the right-hand side an absolutely convergent series for ${|z| < r}$, and the claim follows. $\Box$

Exercise 9 Establish the generalised Cauchy integral formulae

$\displaystyle f^{(k)}(z_0) = \frac{k!}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{k+1}}\ dz$

for any non-negative integer ${k}$, where ${f^{(k)}}$ is the ${k}$-fold complex derivative of ${f}$.

This in turn leads to a converse to Cauchy’s theorem, known as Morera’s theorem:

Corollary 10 (Morera’s theorem) Let ${f: \Omega \rightarrow {\bf C}}$ be a continuous function on an open set ${\Omega}$ with the property that ${\int_\gamma f(z)\ dz = 0}$ for all closed contours ${\gamma: [0,1] \rightarrow \Omega}$. Then ${f}$ is holomorphic.

Proof: We can of course assume ${\Omega}$ to be non-empty and connected (hence path-connected). Fix a point ${z_0 \in \Omega}$, and define a “primitive” ${F: \Omega \rightarrow {\bf C}}$ of ${f}$ by defining ${F(z_1) = \int_\gamma f(z)\ dz}$, with ${\gamma: [0,1] \rightarrow \Omega}$ being any contour from ${z_0}$ to ${z_1}$ (this is well defined by hypothesis). By mimicking the proof of the real fundamental theorem of calculus, we see that ${F}$ is holomorphic with ${F'=f}$, and the claim now follows from Corollary 8. $\Box$

An important consequence of Morera’s theorem for us is

Corollary 11 (Locally uniform limit of holomorphic functions is holomorphic) Let ${f_n: \Omega \rightarrow {\bf C}}$ be holomorphic functions on an open set ${\Omega}$ which converge locally uniformly to a function ${f: \Omega \rightarrow {\bf C}}$. Then ${f}$ is also holomorphic on ${\Omega}$.

Proof: By working locally we may assume that ${\Omega}$ is a ball, and in particular simply connected. By Cauchy’s theorem, ${\int_\gamma f_n(z)\ dz = 0}$ for all closed contours ${\gamma}$ in ${\Omega}$. By local uniform convergence, this implies that ${\int_\gamma f(z)\ dz = 0}$ for all such contours, and the claim then follows from Morera’s theorem. $\Box$

Now we study the zeroes of complex analytic functions. If a complex analytic function ${f}$ vanishes at a point ${z_0}$, but is not identically zero in a neighbourhood of that point, then by Taylor expansion we see that ${f}$ factors in a sufficiently small neighbourhood of ${z_0}$ as

$\displaystyle f(z) = (z-z_0)^n g(z_0) \ \ \ \ \ (4)$

for some natural number ${n}$ (which we call the order or multiplicity of the zero at ${f}$) and some function ${g}$ that is complex analytic and non-zero near ${z_0}$; this generalises the factor theorem for polynomials. In particular, the zero ${z_0}$ is isolated if ${f}$ does not vanish identically near ${z_0}$. We conclude that if ${\Omega}$ is connected and ${f}$ vanishes on a neighbourhood of some point ${z_0}$ in ${\Omega}$, then it must vanish on all of ${\Omega}$ (since the maximal connected neighbourhood of ${z_0}$ in ${\Omega}$ on which ${f}$ vanishes cannot have any boundary point in ${\Omega}$). This implies unique continuation of analytic functions: if two complex analytic functions on ${\Omega}$ agree on a non-empty open set, then they agree everywhere. In particular, if a complex analytic function does not vanish everywhere, then all of its zeroes are isolated, so in particular it has only finitely many zeroes on any given compact set.

Recall that a rational function is a function ${f}$ which is a quotient ${g/h}$ of two polynomials (at least outside of the set where ${h}$ vanishes). Analogously, let us define a meromorphic function on an open set ${\Omega}$ to be a function ${f: \Omega \backslash S \rightarrow {\bf C}}$ defined outside of a discrete subset ${S}$ of ${\Omega}$ (the singularities of ${f}$), which is locally the quotient ${g/h}$ of holomorphic functions, in the sense that for every ${z_0 \in \Omega}$, one has ${f=g/h}$ in a neighbourhood of ${z_0}$ excluding ${S}$, with ${g, h}$ holomorphic near ${z_0}$ and with ${h}$ non-vanishing outside of ${S}$. If ${z_0 \in S}$ and ${g}$ has a zero of equal or higher order than ${h}$ at ${z_0}$, then the singularity is removable and one can extend the meromorphic function holomorphically across ${z_0}$ (by the holomorphic factor theorem (4)); otherwise, the singularity is non-removable and is known as a pole, whose order is equal to the difference between the order of ${h}$ and the order of ${g}$ at ${z_0}$. (If one wished, one could extend meromorphic functions to the poles by embedding ${{\bf C}}$ in the Riemann sphere ${{\bf C} \cup \{\infty\}}$ and mapping each pole to ${\infty}$, but we will not do so here. One could also consider non-meromorphic functions with essential singularities at various points, but we will have no need to analyse such singularities in this course.) If the order of a pole or zero is one, we say that it is simple; if it is two, we say it is double; and so forth.

Exercise 12 Show that the space of meromorphic functions on a non-empty open set ${\Omega}$, quotiented by almost everywhere equivalence, forms a field.

By quotienting two Taylor series, we see that if a meromorphic function ${f}$ has a pole of order ${n}$ at some point ${z_0}$, then it has a Laurent expansion

$\displaystyle f = \sum_{m=-n}^\infty a_m (z-z_0)^m,$

absolutely convergent in a neighbourhood of ${z_0}$ excluding ${z_0}$ itself, and with ${a_{-n}}$ non-zero. The Laurent coefficient ${a_{-1}}$ has a special significance, and is called the residue of the meromorphic function ${f}$ at ${z_0}$, which we will denote as ${\hbox{Res}(f;z_0)}$. The importance of this coefficient comes from the following significant generalisation of the Cauchy integral formula, known as the residue theorem:

Exercise 13 (Residue theorem) Let ${f}$ be a meromorphic function on a simply connected domain ${\Omega}$, and let ${\gamma}$ be a closed contour in ${\Omega}$ enclosing a bounded region ${U}$ anticlockwise, and avoiding all the singularities of ${f}$. Show that

$\displaystyle \int_\gamma f(z)\ dz = 2\pi i \sum_\rho \hbox{Res}(f;\rho)$

where ${\rho}$ is summed over all the poles of ${f}$ that lie in ${U}$.

The residue theorem is particularly useful when applied to logarithmic derivatives ${f'/f}$ of meromorphic functions ${f}$, because the residue is of a specific form:

Exercise 14 Let ${f}$ be a meromorphic function on an open set ${\Omega}$ that does not vanish identically. Show that the only poles of ${f'/f}$ are simple poles (poles of order ${1}$), occurring at the poles and zeroes of ${f}$ (after all removable singularities have been removed). Furthermore, the residue of ${f'/f}$ at a pole ${z_0}$ is an integer, equal to the order of zero of ${f}$ if ${f}$ has a zero at ${z_0}$, or equal to negative the order of pole at ${f}$ if ${f}$ has a pole at ${z_0}$.

Remark 15 The fact that residues of logarithmic derivatives of meromorphic functions are automatically integers is a remarkable feature of the complex analytic approach to multiplicative number theory, which is difficult (though not entirely impossible) to duplicate in other approaches to the subject. Here is a sample application of this integrality, which is challenging to reproduce by non-complex-analytic means: if ${f}$ is meromorphic near ${z_0}$, and one has the bound ${|\frac{f'}{f}(z_0+t)| \leq \frac{0.9}{t} + O(1)}$ as ${t \rightarrow 0^+}$, then ${\frac{f'}{f}}$ must in fact stay bounded near ${z_0}$, because the only integer of magnitude less than ${0.9}$ is zero.

The Riemann zeta function $\zeta(s)$, defined for $\hbox{Re}(s) > 1$ by the formula

$\displaystyle \zeta(s) := \sum_{n \in {\Bbb N}} \frac{1}{n^s}$ (1)

where ${\Bbb N} = \{1,2,\ldots\}$ are the natural numbers, and extended meromorphically to other values of s by analytic continuation, obeys the remarkable functional equation

$\displaystyle \Xi(s) = \Xi(1-s)$ (2)

where

$\displaystyle \Xi(s) := \Gamma_\infty(s) \zeta(s)$ (3)

is the Riemann Xi function,

$\displaystyle \Gamma_\infty(s) := \pi^{-s/2} \Gamma(s/2)$ (4)

is the Gamma factor at infinity, and the Gamma function $\Gamma(s)$ is defined for $\hbox{Re}(s) > 1$ by

$\displaystyle \Gamma(s) := \int_0^\infty e^{-t} t^s\ \frac{dt}{t}$ (5)

and extended meromorphically to other values of s by analytic continuation.

There are many proofs known of the functional equation (2).  One of them (dating back to Riemann himself) relies on the Poisson summation formula

$\displaystyle \sum_{a \in {\Bbb Z}} f_\infty(a t_\infty) = \frac{1}{|t|_\infty} \sum_{a \in {\Bbb Z}} \hat f_\infty(a/t_\infty)$ (6)

for the reals $k_\infty := {\Bbb R}$ and $t \in k_\infty^*$, where $f$ is a Schwartz function, $|t|_\infty := |t|$ is the usual Archimedean absolute value on $k_\infty$, and

$\displaystyle \hat f_\infty(\xi_\infty) := \int_{k_\infty} e_\infty(-x_\infty \xi_\infty) f_\infty(x_\infty)\ dx_\infty$ (7)

is the Fourier transform on $k_\infty$, with $e_\infty(x_\infty) := e^{2\pi i x_\infty}$ being the standard character $e_\infty: k_\infty \to S^1$ on $k_\infty$.  (The reason for this rather strange notation for the real line and its associated structures will be made clearer shortly.)  Applying this formula to the (Archimedean) Gaussian function

$\displaystyle g_\infty(x_\infty) := e^{-\pi |x_\infty|^2}$, (8)

which is its own (additive) Fourier transform, and then applying the multiplicative Fourier transform (i.e. the Mellin transform), one soon obtains (2).  (Riemann also had another proof of the functional equation relying primarily on contour integration, which I will not discuss here.)  One can “clean up” this proof a bit by replacing the Gaussian by a Dirac delta function, although one now has to work formally and “renormalise” by throwing away some infinite terms.  (One can use the theory of distributions to make this latter approach rigorous, but I will not discuss this here.)  Note how this proof combines the additive Fourier transform with the multiplicative Fourier transform.  [Continuing with this theme, the Gamma function (5) is an inner product between an additive character $e^{-t}$ and a multiplicative character $t^s$, and the zeta function (1) can be viewed both additively, as a sum over n, or multiplicatively, as an Euler product.]

In the famous thesis of Tate, the above argument was reinterpreted using the language of the adele ring ${\Bbb A}$, with the Poisson summation formula (4) on $k_\infty$ replaced by the Poisson summation formula

$\displaystyle \sum_{a \in k} f(a t) = \sum_{a \in k} \hat f(t/a)$ (9)

on ${\Bbb A}$, where $k = {\Bbb Q}$ is the rationals, $t \in {\Bbb A}$, and f is now a Schwartz-Bruhat function on ${\Bbb A}$.  Applying this formula to the adelic (or global) Gaussian function $g(x) := g_\infty(x_\infty) \prod_p 1_{{\mathbb Z}_p}(x_p)$, which is its own Fourier transform, and then using the adelic Mellin transform, one again obtains (2).  Again, the proof can be cleaned up by replacing the Gaussian with a Dirac mass, at the cost of making the computations formal (or requiring the theory of distributions).

In this post I will write down both Riemann’s proof and Tate’s proof together (but omitting some technical details), to emphasise the fact that they are, in some sense, the same proof.  However, Tate’s proof gives a high-level clarity to the situation (in particular, explaining more adequately why the Gamma factor at infinity (4) fits seamlessly with the Riemann zeta function (1) to form the Xi function (2)), and allows one to generalise the functional equation relatively painlessly to other zeta-functions and L-functions, such as Dedekind zeta functions and Hecke L-functions.

[Note: the material here is very standard in modern algebraic number theory; the post here is partially for my own benefit, as most treatments of this topic in the literature tend to operate in far higher levels of generality than I would prefer.]