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246B, Notes 1: Zeroes, poles, and factorisation of meromorphic functions

23 December, 2020 in 246B - complex analysis, math.CV | Tags: , , , | by | 74 comments

Previous set of notes: 246A Notes 5. Next set of notes: Notes 2.

— 1. Jensen’s formula —

Suppose {f} is a non-zero rational function {f =P/Q}, then by the fundamental theorem of algebra one can write

\displaystyle f(z) = c \frac{\prod_\rho (z-\rho)}{\prod_\zeta (z-\zeta)}

for some non-zero constant {c}, where {\rho} ranges over the zeroes of {P} (counting multiplicity) and {\zeta} ranges over the zeroes of {Q} (counting multiplicity), and assuming {z} avoids the zeroes of {Q}. Taking absolute values and then logarithms, we arrive at the formula

\displaystyle \log |f(z)| = \log |c| + \sum_\rho \log|z-\rho| - \sum_\zeta \log |z-\zeta|, \ \ \ \ \ (1)

 as long as {z} avoids the zeroes of both {P} and {Q}. (In this set of notes we use {\log} for the natural logarithm when applied to a positive real number, and {\mathrm{Log}} for the standard branch of the complex logarithm (which extends {\log}); the multi-valued complex logarithm {\log} will only be used in passing.) Alternatively, taking logarithmic derivatives, we arrive at the closely related formula

\displaystyle \frac{f'(z)}{f(z)} = \sum_\rho \frac{1}{z-\rho} - \sum_\zeta \frac{1}{z-\zeta}, \ \ \ \ \ (2)

 again for {z} avoiding the zeroes of both {P} and {Q}. Thus we see that the zeroes and poles of a rational function {f} describe the behaviour of that rational function, as well as close relatives of that function such as the log-magnitude {\log|f|} and log-derivative {\frac{f'}{f}}. We have already seen these sorts of formulae arise in our treatment of the argument principle in 246A Notes 4.

Exercise 1 Let {P(z)} be a complex polynomial of degree {n \geq 1}.
  • (i) (Gauss-Lucas theorem) Show that the complex roots of {P'(z)} are contained in the closed convex hull of the complex roots of {P(z)}.
  • (ii) (Laguerre separation theorem) If all the complex roots of {P(z)} are contained in a disk {D(z_0,r)}, and {\zeta \not \in D(z_0,r)}, then all the complex roots of {nP(z) + (\zeta - z) P'(z)} are also contained in {D(z_0,r)}. (Hint: apply a suitable Möbius transformation to move {\zeta} to infinity, and then apply part (i) to a polynomial that emerges after applying this transformation.)

There are a number of useful ways to extend these formulae to more general meromorphic functions than rational functions. Firstly there is a very handy “local” variant of (1) known as Jensen’s formula:

Theorem 2 (Jensen’s formula) Let {f} be a meromorphic function on an open neighbourhood of a disk {\overline{D(z_0,r)} = \{ z: |z-z_0| \leq r \}}, with all removable singularities removed. Then, if {z_0} is neither a zero nor a pole of {f}, we have

\displaystyle \log |f(z_0)| = \int_0^1 \log |f(z_0+re^{2\pi i t})|\ dt + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{|\rho-z_0|}{r} \ \ \ \ \ (3)

\displaystyle - \sum_{\zeta: |\zeta-z_0| \leq r} \log \frac{|\zeta-z_0|}{r}

where {\rho} and {\zeta} range over the zeroes and poles of {f} respectively (counting multiplicity) in the disk {\overline{D(z_0,r)}}.

One can view (3) as a truncated (or localised) variant of (1). Note also that the summands {\log \frac{|\rho-z_0|}{r}, \log \frac{|\zeta-z_0|}{r}} are always non-positive.

Proof: By perturbing {r} slightly if necessary, we may assume that none of the zeroes or poles of {f} (which form a discrete set) lie on the boundary circle {\{ z: |z-z_0| = r \}}. By translating and rescaling, we may then normalise {z_0=0} and {r=1}, thus our task is now to show that

\displaystyle \log |f(0)| = \int_0^1 \log |f(e^{2\pi i t})|\ dt + \sum_{\rho: |\rho| < 1} \log |\rho| - \sum_{\zeta: |\zeta| < 1} \log |\zeta|. \ \ \ \ \ (4)

We may remove the poles and zeroes inside the disk {D(0,1)} by the useful device of Blaschke products. Suppose for instance that {f} has a zero {\rho} inside the disk {D(0,1)}. Observe that the function

\displaystyle B_\rho(z) := \frac{\rho - z}{1 - \overline{\rho} z} \ \ \ \ \ (5)

 has magnitude {1} on the unit circle {\{ z: |z| = 1\}}, equals {\rho} at the origin, has a simple zero at {\rho}, but has no other zeroes or poles inside the disk. Thus Jensen’s formula (4) already holds if {f} is replaced by {B_\rho}. To prove (4) for {f}, it thus suffices to prove it for {f/B_\rho}, which effectively deletes a zero {\rho} inside the disk {D(0,1)} from {f} (and replaces it instead with its inversion {1/\overline{\rho}}). Similarly we may remove all the poles inside the disk. As a meromorphic function only has finitely many poles and zeroes inside a compact set, we may thus reduce to the case when {f} has no poles or zeroes on or inside the disk {D(0,1)}, at which point our goal is simply to show that

\displaystyle \log |f(0)| = \int_0^1 \log |f(e^{2\pi i t})|\ dt.

Since {f} has no zeroes or poles inside the disk, it has a holomorphic logarithm {F} (Exercise 46 of 246A Notes 4). In particular, {\log |f|} is the real part of {F}. The claim now follows by applying the mean value property (Exercise 17 of 246A Notes 3) to {\log |f|}. \Box

An important special case of Jensen’s formula arises when {f} is holomorphic in a neighborhood of {\overline{D(z_0,r)}}, in which case there are no contributions from poles and one simply has

\displaystyle \int_0^1 \log |f(z_0+re^{2\pi i t})|\ dt = \log |f(z_0)| + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{r}{|\rho-z_0|}. \ \ \ \ \ (6)

This is quite a useful formula, mainly because the summands {\log \frac{r}{|\rho-z_0|}} are non-negative; it can be viewed as a more precise assertion of the subharmonicity of {\log |f|} (see Exercises 60(ix) and 61 of 246A Notes 5). Here are some quick applications of this formula:

Exercise 3 Use (6) to give another proof of Liouville’s theorem: a bounded holomorphic function {f} on the entire complex plane is necessarily constant.

Exercise 4 Use Jensen’s formula to prove the fundamental theorem of algebra: a complex polynomial {P(z)} of degree {n} has exactly {n} complex zeroes (counting multiplicity), and can thus be factored as {P(z) = c (z-z_1) \dots (z-z_n)} for some complex numbers {c,z_1,\dots,z_n} with {c \neq 0}. (Note that the fundamental theorem was invoked previously in this section, but only for motivational purposes, so the proof here is non-circular.)

Exercise 5 (Shifted Jensen’s formula) Let {f} be a meromorphic function on an open neighbourhood of a disk {\{ z: |z-z_0| \leq r \}}, with all removable singularities removed. Show that

\displaystyle \log |f(z)| = \int_0^1 \log |f(z_0+re^{2\pi i t})| \mathrm{Re} \frac{r e^{2\pi i t} + (z-z_0)}{r e^{2\pi i t} - (z-z_0)}\ dt \ \ \ \ \ (7)

\displaystyle + \sum_{\rho: |\rho-z_0| \leq r} \log \frac{|\rho-z|}{|r - \rho^* (z-z_0)|}

\displaystyle - \sum_{\zeta: |\zeta-z_0| \leq r} \log \frac{|\zeta-z|}{|r - \zeta^* (z-z_0)|}

for all {z} in the open disk {\{ z: |z-z_0| < r\}} that are not zeroes or poles of {f}, where {\rho^* = \frac{\overline{\rho-z_0}}{r}} and {\zeta^* = \frac{\overline{\zeta-z_0}}{r}}. (The function {\Re \frac{r e^{2\pi i t} + (z-z_0)}{r e^{2\pi i t} - (z-z_0)}} appearing in the integrand is sometimes known as the Poisson kernel, particularly if one normalises so that {z_0=0} and {r=1}.)

Exercise 6 (Bounded type)
  • (i) If {f} is a holomorphic function on {D(0,1)} that is not identically zero, show that {\liminf_{r \rightarrow 1^-} \int_0^{2\pi} \log |f(re^{i\theta})|\ d\theta > -\infty}.
  • (ii) If {f} is a meromorphic function on {D(0,1)} that is the ratio of two bounded holomorphic functions that are not identically zero, show that {\limsup_{r \rightarrow 1^-} \int_0^{2\pi} |\log |f(re^{i\theta})||\ d\theta < \infty}. (Functions {f} of this form are said to be of bounded type and lie in the Nevanlinna class for the unit disk {D(0,1)}.)

Exercise 7 (Smoothed out Jensen formula) Let {f} be a meromorphic function on an open set {U}, and let {\phi: U \rightarrow {\bf C}} be a smooth compactly supported function. Show that

\displaystyle \sum_\rho \phi(\rho) - \sum_\zeta \phi(\zeta)

\displaystyle = \frac{-1}{2\pi} \int\int_U ((\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}) \phi(x+iy)) \frac{f'}{f}(x+iy)\ dx dy

\displaystyle = \frac{1}{2\pi} \int\int_U ((\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}) \phi(x+iy)) \log |f(x+iy)|\ dx dy

where {\rho, \zeta} range over the zeroes and poles of {f} (respectively) in the support of {\phi}. Informally argue why this identity is consistent with Jensen’s formula. (Note: as many of the functions involved here are not holomorphic, complex analysis tools are of limited use. Try using real variable tools such as Stokes theorem, Greens theorem, or integration by parts.)

When applied to entire functions {f}, Jensen’s formula relates the order of growth of {f} near infinity with the density of zeroes of {f}. Here is a typical result:

Proposition 8 Let {f: {\bf C} \rightarrow {\bf C}} be an entire function, not identically zero, that obeys a growth bound {|f(z)| \leq C \exp( C|z|^\alpha)} for some {C, \alpha > 0} and all {z}. Then there exists a constant {C'>0} such that {D(0,R)} has at most {C' R^\alpha} zeroes (counting multiplicity) for any {R \geq 1}.

Entire functions that obey a growth bound of the form {|f(z)| \leq C_\varepsilon \exp( C_\varepsilon |z|^{\rho+\varepsilon})} for every {\varepsilon>0} and {z} (where {C_\varepsilon} depends on {\varepsilon}) are said to be of order at most {\rho}. The above theorem shows that for such functions that are not identically zero, the number of zeroes in a disk of radius {R} does not grow much faster than {R^\rho}. This is often a useful preliminary upper bound on the zeroes of entire functions, as the order of an entire function tends to be relatively easy to compute in practice.

Proof: First suppose that {f(0)} is non-zero. From (6) applied with {r=2R} and {z_0=0} one has

\displaystyle \int_0^1 \log(C \exp( C (2R)^\alpha ) )\ dt \geq \log |f(0)| + \sum_{\rho: |\rho| \leq 2R} \log \frac{2R}{|\rho|}.

Every zero in {D(0,R)} contribute at least {\log 2} to a summand on the right-hand side, while all other zeroes contribute a non-negative quantity, thus

\displaystyle \log C + C (2R)^\alpha \geq \log |f(0)| + N_R \log 2

where {N_R} denotes the number of zeroes in {D(0,R)}. This gives the claim for {f(0) \neq 0}. When {f(0)=0}, one can shift {f} by a small amount to make {f} non-zero at the origin (using the fact that zeroes of holomorphic functions not identically zero are isolated), modifying {C} in the process, and then repeating the previous arguments. \Box

Just as (3) and (7) give truncated variants of (1), we can create truncated versions of (2). The following crude truncation is adequate for many applications:

Theorem 9 (Truncated formula for log-derivative) Let {f} be a holomorphic function on an open neighbourhood of a disk {\{ z: |z-z_0| \leq r \}} that is not identically zero on this disk. Suppose that one has a bound of the form {|f(z)| \leq M^{O_{c_1,c_2}(1)} |f(z_0)|} for some {M \geq 1} and all {z} on the circle {\{ z: |z-z_0| = r\}}. Let {0 < c_2 < c_1 < 1} be constants. Then one has the approximate formula

\displaystyle \frac{f'(z)}{f(z)} = \sum_{\rho: |\rho - z_0| \leq c_1 r} \frac{1}{z-\rho} + O_{c_1,c_2}( \frac{\log M}{r} )

for all {z} in the disk {\{ z: |z-z_0| < c_2 r \}} other than zeroes of {f}. Furthermore, the number of zeroes {\rho} in the above sum is {O_{c_1,c_2}(\log M)}.

Proof: To abbreviate notation, we allow all implied constants in this proof to depend on {c_1,c_2}.

We mimic the proof of Jensen’s formula. Firstly, we may translate and rescale so that {z_0=0} and {r=1}, so we have {|f(z)| \leq M^{O(1)} |f(0)|} when {|z|=1}, and our main task is to show that

\displaystyle \frac{f'(z)}{f(z)} - \sum_{\rho: |\rho| \leq c_1} \frac{1}{z-\rho} = O( \log M ) \ \ \ \ \ (8)

for {|z| \leq c_2}. Note that if {f(0)=0} then {f} vanishes on the unit circle and hence (by the maximum principle) vanishes identically on the disk, a contradiction, so we may assume {f(0) \neq 0}. From hypothesis we then have

\displaystyle \log |f(z)| \leq \log |f(0)| + O(\log M)

on the unit circle, and so from Jensen’s formula (3) we see that

\displaystyle \sum_{\rho: |\rho| \leq 1} \log \frac{1}{|\rho|} = O(\log M). \ \ \ \ \ (9)

 In particular we see that the number of zeroes with {|\rho| \leq c_1} is {O(\log M)}, as claimed.

Suppose {f} has a zero {\rho} with {c_1 < |\rho| \leq 1}. If we factor {f = B_\rho g}, where {B_\rho} is the Blaschke product (5), then

\displaystyle \frac{f'}{f} = \frac{B'_\rho}{B_\rho} + \frac{g'}{g}

\displaystyle = \frac{g'}{g} + \frac{1}{z-\rho} - \frac{1}{z-1/\overline{\rho}}.

Observe from Taylor expansion that the distance between {\rho} and {1/\overline{\rho}} is {O( \log \frac{1}{|\rho|} )}, and hence {\frac{1}{z-\rho} - \frac{1}{z-1/\overline{\rho}} = O( \log \frac{1}{|\rho|} )} for {|z| \leq c_2}. Thus we see from (9) that we may use Blaschke products to remove all the zeroes in the annulus {c_1 < |\rho| \leq 1} while only affecting the left-hand side of (8) by {O( \log M)}; also, removing the Blaschke products does not affect {|f(z)|} on the unit circle, and only affects {\log |f(0)|} by {O(\log M)} thanks to (9). Thus we may assume without loss of generality that there are no zeroes in this annulus.

Similarly, given a zero {\rho} with {|\rho| \leq c_1}, we have {\frac{1}{z-1/\overline{\rho}} = O(1)}, so using Blaschke products to remove all of these zeroes also only affects the left-hand side of (8) by {O(\log M)} (since the number of zeroes here is {O(\log M)}), with {\log |f(0)|} also modified by at most {O(\log M)}. Thus we may assume in fact that {f} has no zeroes whatsoever within the unit disk. We may then also normalise {f(0) = 1}, then {\log |f(e^{2\pi i t})| \leq O(\log M)} for all {t \in [0,1]}. By Jensen’s formula again, we have

\displaystyle \int_0^1 \log |f(e^{2\pi i t})|\ dt = 0

and thus (by using the identity {|x| = 2 \max(x,0) - x} for any real {x})

\displaystyle \int_0^1 \left|\log |f(e^{2\pi i t})|\right|\ dt \ll \log M. \ \ \ \ \ (10)

 On the other hand, from (7) we have

\displaystyle \log |f(z)| = \int_0^1 \log |f(e^{2\pi i t})| \mathrm{Re} \frac{e^{2\pi i t} + z}{e^{2\pi i t} - z}\ dt

which implies from (10) that {\log |f(z)|} and its first derivatives are {O( \log M )} on the disk {\{ z: |z| \leq c_2 \}}. But recall from the proof of Jensen’s formula that {\frac{f'}{f}} is the derivative of a logarithm {\log f} of {f}, whose real part is {\log |f|}. By the Cauchy-Riemann equations for {\log f}, we conclude that {\frac{f'}{f} = O(\log M)} on the disk {\{ z: |z| \leq c_2 \}}, as required. \Box

Exercise 10
  • (i) (Borel-Carathéodory theorem) If {f: U \rightarrow {\bf C}} is analytic on an open neighborhood of a disk {\overline{D(z_0,R)}} and {0 < r < R}, show that

    \displaystyle \sup_{z \in D(z_0,r)} |f(z)| \leq \frac{2r}{R-r} \sup_{z \in \overline{D(z_0,R)}} \mathrm{Re} f(z) + \frac{R+r}{R-r} |f(z_0)|.

    (Hint: one can normalise {z_0=0}, {R=1}, {f(0)=0}, and {\sup_{|z-z_0| \leq R} \mathrm{Re} f(z)=1}. Now {f} maps the unit disk to the half-plane {\{ \mathrm{Re} z \leq 1 \}}. Use a Möbius transformation to map the half-plane to the unit disk and then use the Schwarz lemma.)
  • (ii) Use (i) to give an alternate way to conclude the proof of Theorem 9.

A variant of the above argument allows one to make precise the heuristic that holomorphic functions locally look like polynomials:

Exercise 11 (Local Weierstrass factorisation) Let the notation and hypotheses be as in Theorem 9. Then show that

\displaystyle f(z) = P(z) \exp( g(z) )

for all {z} in the disk {\{ z: |z-z_0| < c_2 r \}}, where {P} is a polynomial whose zeroes are precisely the zeroes of {f} in {\{ z: |z-z_0| \leq c_1r \}} (counting multiplicity) and {g} is a holomorphic function on {\{ z: |z-z_0| < c_2 r \}} of magnitude {O_{c_1,c_2}( \log M )} and first derivative {O_{c_1,c_2}( \log M / r )} on this disk. Furthermore, show that the degree of {P} is {O_{c_1,c_2}(\log M)}.

Exercise 12 (Preliminary Beurling factorisation) Let {H^\infty(D(0,1))} denote the space of bounded analytic functions {f: D(0,1) \rightarrow {\bf C}} on the unit disk; this is a normed vector space with norm

\displaystyle \|f\|_{H^\infty(D(0,1))} := \sup_{z \in D(0,1)} |f(z)|.

  • (i) If {f \in H^\infty(D(0,1))} is not identically zero, and {z_n} denote the zeroes of {f} in {D(0,1)} counting multiplicity, show that

    \displaystyle \sum_n (1-|z_n|) < \infty

    and

    \displaystyle \sup_{1/2 < r < 1} \int_0^{2\pi} | \log |f(re^{i\theta})| |\ d\theta < \infty.

  • (ii) Let the notation be as in (i). If we define the Blaschke product

    \displaystyle B(z) := z^m \prod_{|z_n| \neq 0} \frac{|z_n|}{z_n} \frac{z_n-z}{1-\overline{z_n} z}

    where {m} is the order of vanishing of {f} at zero, show that this product converges absolutely to a holomorphic function on {D(0,1)}, and that {|f(z)| \leq \|f\|_{H^\infty(D(0,1)} |B(z)|} for all {z \in D(0,1)}. (It may be easier to work with finite Blaschke products first to obtain this bound.)
  • (iii) Continuing the notation from (i), establish a factorisation {f(z) = B(z) \exp(g(z))} for some holomorphic function {g: D(0,1) \rightarrow {\bf C}} with {\mathrm{Re}(g(z)) \leq \log \|f\|_{H^\infty(D(0,1)}} for all {z\in D(0,1)}.
  • (iv) (Theorem of F. and M. Riesz, special case) If {f \in H^\infty(D(0,1))} extends continuously to the boundary {\{e^{i\theta}: 0 \leq \theta < 2\pi\}}, show that the set {\{ 0 \leq \theta < 2\pi: f(e^{i\theta})=0 \}} has zero measure.

Remark 13 The factorisation (iii) can be refined further, with {g} being the Poisson integral of some finite measure on the unit circle. Using the Lebesgue decomposition of this finite measure into absolutely continuous parts one ends up factorising {H^\infty(D(0,1))} functions into “outer functions” and “inner functions”, giving the Beurling factorisation of {H^\infty}. There are also extensions to larger spaces {H^p(D(0,1))} than {H^\infty(D(0,1))} (which are to {H^\infty} as {L^p} is to {L^\infty}), known as Hardy spaces. We will not discuss this topic further here, but see for instance this text of Garnett for a treatment.

Exercise 14 (Littlewood’s lemma) Let {f} be holomorphic on an open neighbourhood of a rectangle {R = \{ \sigma+it: \sigma_0 \leq \sigma \leq \sigma_1; 0 \leq t \leq T \}} for some {\sigma_0 < \sigma_1} and {T>0}, with {f} non-vanishing on the boundary of the rectangle. Show that

\displaystyle 2\pi \sum_\rho (\mathrm{Re}(\rho)-\sigma_0) = \int_0^T \log |f(\sigma_0+it)|\ dt - \int_0^T \log |f(\sigma_1+it)|\ dt

\displaystyle + \int_{\sigma_0}^{\sigma_1} \mathrm{arg} f(\sigma+iT)\ d\sigma - \int_{\sigma_0}^{\sigma_1} \mathrm{arg} f(\sigma)\ d\sigma

where {\rho} ranges over the zeroes of {f} inside {R} (counting multiplicity) and one uses a branch of {\mathrm{arg} f} which is continuous on the upper, lower, and right edges of {C}. (This lemma is a popular tool to explore the zeroes of Dirichlet series such as the Riemann zeta function.)

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254A, Supplement 2: A little bit of complex and Fourier analysis

5 December, 2014 in 254A - analytic prime number theory, math.CA, math.CV | Tags: , , , , , | by | 24 comments

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.

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