In this section, we shall demonstrate how far an analytic function may be characterized by the nature of its singularities and roots, a singular point being a place where the regularity property fails. In particular, we are concerned with functions whose singularities are isolated. A singular point \(z_{0}\) is said to be an isolated singularity of a single-valued function \(f(z)\) if it is completely surrounded by regular points; i.e., if \(f(z)\) is regular in some deleted neighborhood \(0<|z-z_{0}|<\varepsilon\) .

4.2.1 The Nature of Singular Points

The most natural instances of isolated singularities that come to mind are like the singularities of the functions \(1 / z\) and \(e^{1 / z}\) at the origin \(z=0\) . Another example is given by the function defined to be \(1\) at the origin and zero everywhere else. Note, however, that the point \({z}=0\) is not an isolated singularity for the multi-valued function \(f(z)=\sqrt{z}\) . True, \(\sqrt{z}\) does not possess a derivative at the origin, but there is no neighborhood of the origin in which the function is single-valued. In other words, the requirement that \(f(z)\) be single-valued is meant to exclude the consideration of branch points.

We have at hand a convenient tool, the Laurent series, for analyzing the behavior of a function at an isolated singularity. Set \[f(z)=\sum_{-\infty}^{+\infty} a_{n}(z-z_{0})^{n}.\] The series converges in any sufficiently small circle about \({z}_{0}\) .

We divide isolated singularities into three classes according to the local behavior of the function.

First of all \(f(z)\) may be bounded in some neighborhood of \(z_{0}\) , \[|f(z)| \leq M .\] In that case we have (after (1.23) ) \(|a_{n}| \leq M / \rho^{n}\) where \(\rho\) can be made arbitrarily small. We conclude that all the coefficients with negative index vanish. Hence, except for the point \(z_{0}\) , \(f(z)\) may be represented by a simple Taylor series \[f(z)=a_{0}+a_{1}(z-z_{0})+a_{2}(z-z_{0})^2+\cdots.\] The Taylor series defines an analytic function throughout the interior of a circle about \(z_{0}\) and this function coincides with \(f(z)\) at all of these points but the center \(z=z_{0}\) . Hence if we redefine \(f(z)\) at \(z_{0}\) by setting \(f(z_{0})=a_{0}\) then the function will be regular. In other words, if \(f(z)\) is regular and \(|f(z)|\) is bounded in the deleted neighborhood of \(z_{0}\) , then the function will have a limiting value at \({z}_{0}\) . The singularity can only be due to a break in the continuity of \(f(z)\) . Thus, in order to make the function regular at \(z_{0}\) it suffices to redefine \(f(z)\) at the one point \(z_{0}\) so as to make \(f(z)\) continuous. For this reason, singularities of this type are said to be removable .

On the other hand, if \(f(z)\) is not bounded in the neighborhood of \(z_{0}\) then terms of negative index will appear in the Laurent expansion. We may then write \[f(z)=g(z-z_{0})+h\Big(\frac{1}{z-z_{0}}\Big)\] where \begin{align} g(z-z_{0})&=\sum_{n=0}^{\infty} a_{n}(z-z_{0})^{n} ;\\ h\Big(\frac{1}{z-z_{0}}\Big)&=\sum_{n=1}^{\infty} a_{-n}\Big(\frac{1}{z-z_{0}}\Big)^{n}. \end{align} The series \(h\big(\frac{1}{z-z_{0}}\big)\) of powers of \(\frac{1}{z-z_{0}}\) is called the principal part of \(f(z)\) . Now, if the principal part of \(f(z)\) consists of only a finite number of terms; \(a_{-n}=0\) for \(n>k>0\) and \(a_{k} \neq 0\) , then we say that \[f(z)=\frac{a_{-k}}{(z-z_{0})^{k}}+\frac{a_{-k+1}}{(z-z_{0})^{k-1}}+\cdots\] has a pole of order \(\boldsymbol{k}\) at \(z=z_{0}\) . If, on the contrary, the principal part of the Laurent series has an infinite number of terms we say that \(z_{0}\) is an essential singularity of \(f(z)\) . The function \(e^{1 / z}\) exhibits such a singularity at \(z=0\) .

If \(f(z)\) has a pole of order \(n\) at \(z_{0}\) then clearly \[(z-z_{0})^{n} f(z)\] is regular at \(z_{0}\) . Conversely, if \((z-z_{0})^{n} f(z)\) is regular at \(z_{0}\) then \(f(z)\) can have at most a pole of \(n^\text{th}\) order at \(z_{0}\) . Clearly, if \(f(z)\) is regular and possesses a zero of \(n^\text{th}\) order at \(z_{0}\) then \(\frac{1}{f(z)}\) has a pole of \(n^\text{th}\) order at \(z_{0}\) .

At a pole \(z_{0}\) the function \(f(z)\) becomes infinite: more precisely, given any \({M}\) , however large, there is an entire neighborhood \(0 < |z-z_{0}| \leq \varepsilon\) in which \[|f(z)| \geq M.\] This is immediately clear from the fact that \(f(z)\) can be written in the form \[f(z)=\frac{g(z)}{(z-z_{0})^{k}}\] where \(g(z)\) is regular in the neighborhood of \(z_{0}\) .

In contrast to the behavior of a function in the neighborhood of a pole we have the Theorem of Weierstrass .

There is a much stronger result due to Picard which states that, in every neighborhood of an essential singularity, an analytic function actually takes on every complex value, excepting at most two. We shall prove this result in a later chapter.

We broaden the definition of essential singularity to include all singularities (not necessarily isolated) which are not poles or removable singularities. Hence a point of accumulation of poles in a domain where \(f(z)\) is otherwise regular is an essential singularity. The nomenclature is justified, for an accumulation point \(z_{0}\) of poles does not behave like a pole. Suppose that \(f(z)\) could be bounded below in a neighborhood of \(z_{0}\) , \(|f(z)| \geq M\) . Then in that neighborhood the function \(\frac{1}{f(z)}\) would be regular and \(z_{0}\) would be an accumulation point of zeros for \(\frac{1}{f(z)}\) in its domain of regularity. But this would imply that \(1 / f(z)=0\) identically and that is clearly impossible.

The Point \(\boldsymbol{\infty}\) 

We shall adopt certain conventions for describing the behavior of an analytic function at infinity. Suppose that \(f(z)\) is regular in the exterior of some circle. We set \(\zeta=\frac{1}{z}\) and define \[g(\zeta)=f\Big(\frac{1}{\zeta}\Big)\] \({g}(\zeta)\) is evidently a regular function in the neighborhood of the origin except perhaps at \(\zeta=0\) . We ascribe to \(f(z)\) at \(z=\infty\) the behavior of the function \(g(\zeta)\) at the origin.

If \(f(z)\) is bounded for sufficiently large values of \(|z|\) then \(g(\zeta)\) will be bounded in the neighborhood of \(\zeta=0\) and therefore can be completed to be regular at the origin. We then say that \(\infty\) is a regular point of \(f(z)\) . If \(f(z)\) is regular at infinity then from \[f'(z)=-\zeta^{2} g'(\zeta)\] we conclude that the derivative of \(f(z)\) must have a zero of at least second order at \(\infty\) .

If \(f(z)\) is regular at infinity it may easily be represented by a Laurent series expanded about the origin. In fact, we have the Taylor expansion for \({g}(\zeta)\) : \[g(\zeta)=a_{0}+a_{1} \zeta+a_{2} \zeta^{2}+\cdots\] whence \[f(z)=a_{0}+\frac{a_{1}}{z}+\frac{a_{2}}{z^{2}}+\cdots.\] 

Exercises

4.2.2 The Zeros and Poles of an Analytic Function

Let \(f(z) \not\equiv 0\) be a regular function with the possible exception of poles in a simple connected domain \(D\) . Clearly, there can be no accumulation point of either poles or zeros in \(D\) , for in the first case \(f(z)\) would have an essential singularity, in the second \(f(z)\) would vanish identically. Now, let \(C\) be a simple closed curve in \(D\) which avoids the poles and zeros of the function. The integral \[\frac{1}{2 \pi i} \oint \frac{f'(z)}{f(z)} \,d z\] is equal to the sum of the residues of \(f'(z) / f(z)\) at its singularities inside \(C\) . These singularities, as we now show, will occur only at the zeros and poles of \(f(z)\) .

Suppose that \(a\) is a zero of \({n}^\text{th}\) order. Then in a neighborhood of \(a\) we may set \[f(z)=(z-a)^{n} g(z)\] where \(g(z)\) is regular and \(g(a) \neq 0\) . Hence \[f'(z)=n(z-a)^{n-1} g(z)+(z-a)^{n} g'(z).\] Therefore \[\frac{f'(z)}{f(z)}=\frac{n}{z-a}+\frac{g'(z)}{g(z)};\] here \(g'(z) / g(z)\) is perfectly regular at \(z=a\) . The residue at \(z=a\) , either by (1.24) or directly from the definition, is equal to \(n\) .

Similarly, if \(f(z)\) has a pole at \(z=a\) then in the neighborhood of \(a\) \[f(z)=(z-a)^{-m} h(z)\] where \(h(z)\) is regular and \(h(a) \neq 0\) . It follows as before that \(f'(z) / f(z)\) has a simple pole at \(z=a\) with residue \(-m\) .

We conclude that \[\tag{2.21} \frac{1}{2 \pi i} \int_{C} \frac{f'(z)}{f(z)} \,d z=Z-P\] where \({Z}\) and \({P}\) denote respectively the number of zeros and the number of poles of \(f(z)\) inside \(C\) , each being counted with its order.

If \(f(z)\) is regular in the full plane with the exception only of poles we obtain an interesting consequence. The integral (2.21) taken in the negative sense may validly be interpreted as an integral in the positive sense about the point \(\infty\) . ¹ Thus the integral in reverse orientation gives us the number \(Z_{e}-P_{e}\) ; here \(Z_{e}\) is the total count of zeros, \(P_{e}\) the total count of poles in the exterior of \(C\) . Hence \(Z_{e}-P_{e}=P-Z\) . In other words, for a function with no essential singularity in the full plane the count of zeros is equal to the count of poles.

Since we have a means of counting the zeros and poles of an analytic function we can give a proof of the fundamental theorem of algebra which is more satisfying than that of Chapter III since it gives a complete count of the roots. We prove that a polynomial of \(n^\text{th}\) degree \[f(z)=a_{0}+a_{1} z+\cdots+a_{n} z^{n} \quad (a_{n} \neq 0)\] has precisely \(n\) zeros. The only singularity of \(f(z)\) is a pole of order \(n\) at \(\infty\) . Since the count of roots is the same as the count of poles the number of zeros of \(f(z)\) counted with their multiplicities is exactly \(n\) .

The integral (2.21) has an interesting geometrical meaning. We have \[\tag{2.22} \int_{z_1}^{z_2} \frac{f'(z)}{f(z)} \,d z=\log f(z_{2})-\log f(z_{1}).\] Now, the function \(\log w\) increases by \(2 \pi i\) if \(w\) describes a closed circuit counterclockwise about the origin and decreases by \(2 \pi i\) if the circuit is described in the opposite sense. Hence (2.21) counts the number of times the point \(w=f(z)\) encircles the origin in the \(w\) -plane as \(z\) describes the curve \(C\) in the \(z\) -plane once in the positive sense. Since \[\log f(z_{1})-\log f(z_{2})=\log \left|\frac{f(z_{1})}{f(z_{2})}\right|+i\left[\operatorname{am}\big(f(z_{1})\big)-\operatorname{am} \big(f(z_{2})\big)\right]\] and \(\log |w|\) is single-valued we obtain \[\tag{2.23} Z-P=\frac{1}{2 \pi i} \int_{C} \frac{f'(z)}{f(z)} \,d z=\frac{1}{2 \pi} \Delta_{C} \operatorname{am}\big(f(z)\big)\] where \(\Delta_{C} \operatorname{am}\big(f(z)\big)\) denotes the total change in \(\operatorname{am}\big(f(z)\big)\) as \(z\) traverses the closed path \(C\) . In particular, if \(f(z)\) is regular in \(D\) we obtain for the number of zeros \[\tag{2.24} 2 \pi Z=\Delta_{C} \operatorname{am}\big(f(z)\big).\] 

A useful criterion for the comparison of the zeros of two functions is given by Rouché’s Theorem .

Rouché’s Theorem provides us with another proof of the fundamental theorem of algebra. The polynomial \(f(z)=a_{n} z^{n}\) has \(n\) zeros and on a sufficiently large circle satisfies the conditions of the theorem with respect to \(g(z)=a_{n-1}z^{n-1}+\cdots+a_{0}\) .

Exercises

4.2.3 Entire and Meromorphic Functions

An important class of analytic functions consists of those functions which have no essential singularity at any finite point of the plane. These we divide into several categories.

An entire function is a function which is regular throughout the finite part of the plane. Unless an entire function is constant it must have a singularity at infinity; if the singularity is a pole the function is said to be an entire rational function ; if \({z}=\infty\) is an essential singularity the function is called an entire transcendental function .

A meromorphic function has no finite singularities except poles. Again, it is said to be rational meromorphic if \({z}=\infty\) is a regular point or a pole, transcendental meromorphic if \({z}=\infty\) is an essential singularity.

The polynomials are examples of entire rational functions. Conversely, we now prove that an entire rational function can be only a polynomial. For, if \(f(z)\) has a pole at \(z=\infty\) we may write \[\tag{2.30} f(z)=a_{n} z^{n}+\cdots+a_{0}+\frac{a_{-1}}{z}+\frac{a_{-2}}{z^{2}}+\cdots\] where there are only a finite number of terms of positive index. The polynomial \[\phi(z)=a_{n} z^{n}+\cdots+a_{0}\] coincides with \(f(z)\) at infinity. Since \(f(z)\) and \(\phi(z)\) are regular everywhere else in the plane the function \(f(z)-\phi(z)\) must be regular in the full plane including the point infinity, therefore uniformly bounded. Hence, by Liouville’s Theorem , \[f(z)=\phi(z)+\textit{constant}.\] 

In the same fashion we show that a rational meromorphic function must be a rational function, the quotient of two polynomials. For let \(f(z)\) be a rational meromorphic function and suppose that the points \(z_{0}, \ldots, z_{n}, \infty\) include all the poles of \(f(z)\) . The number of poles is finite since otherwise they would possess a point of accumulation in the full plane and the function would possess an essential singularity. Let \(p_{0}\big(\frac{1}{z-z_{0}}\big), \ldots, p_{n}\big(\frac{1}{z-z_{n}}\big), p_{\infty}(z)\) denote the principal parts of the Laurent expansions of \(f(z)\) about the point \(z_{0}, \ldots, z_{n}, \infty\) respectively. By the same reasoning as before the function \[f(z)-\left[p_{\infty}(z)+\sum_{\nu=0}^{n} p\Big(\frac{1}{z-z_{\nu}}\Big)\right]\] is regular in the full \({z}\) -plane and hence \[\tag{2.31} f(z)=p_{\infty}(z)+\sum_{\nu=0}^{\infty} p\Big(\frac{1}{z - z_\nu}\Big) + \textit{constant}.\] But the \(p_{i}\) are polynomials in their respective arguments. It follows that \(f(z)\) is a rational function, the quotient of two polynomials, \[f(z)=\frac{P(z)}{Q(z)}\] and that (2.31) gives the decomposition of \(f(z)\) into partial fractions, the \(z_{\nu}\) being the zeros of \(Q(z)\) .

Exercises

An entire rational function, that is, a polynomial, can be represented to within a constant multiple as a product of linear factors. For suppose \(z_{1}, z_{2}, \ldots, z_{n}\) are the roots of a polynomial \(f(z)\) of \(n^\text{th}\) degree. Then we have \[\frac{f(z)}{(z-z_{1})(z-z_{2}) \ldots(z-z_{n})},\] a regular function in the full \({z}\) -plane and hence constant. Thus, given the roots of a polynomial \(f(z)\) we may specify \(f(z)\) to within a multiplicative constant \[f(z)=\alpha(z-z_{1}) \ldots(z-z_{n}).\] We are led to ask more generally, how completely do the roots characterize an entire (not necessarily rational) function?

Again, in (2.31) we saw that a rational meromorphic function is determined to within an additive constant by its behavior at poles. How completely is an arbitrary meromorphic function defined by the character of its poles?

The answers to these questions are given in two remarkable theorems of Weierstrass and Mittag-Leffler; the one gives an infinite product representation for entire functions, the other represents a meromorphic function by an infinite decomposition into partial fractions.

Before we prove these theorems let us first indicate the limitations on these results. Suppose \(\phi(z)\) and \(\psi(z)\) are entire functions with the same zeros. The function \(\chi(z)=\phi(z) / \psi(z)\) has no zeros and is also an entire function. As an exercise the reader may verify that \(\chi(z)\) can be written in the form \(\chi(z)=e^{\omega(z)}\) where \(\omega(z)\) is an entire function. It follows that an entire function is determined by its zeros only to within a factor \(e^{\omega(z)}\) where \(\omega(z)\) is a completely arbitrary entire function. For a meromorphic function it may just as easily be shown that it is characterized by the nature of its poles only to within an added entire function.

In effect we have already answered our question for meromorphic functions when the number of poles is finite. The only other possibility is that \(\infty\) is an accumulation point of poles, for obviously no finite accumulation point of poles may be admitted. The complete answer is then given by the Mittag-Leffler Theorem .

In general, to ensure convergence of the series (2.32) , the degree of the polynomials \(q_{\nu}(z)\) will not be uniformly bounded. However, in special circumstances, all the \(q_{\nu}(z)\) may be chosen of the same finite degree. In particular, if the function has only simple poles \[p_{\nu}\Big(\frac{1}{z-z_{\nu}}\Big)=\frac{\alpha_{\nu}}{z-z_{\nu}}\] then the \(q_{\nu}(z)\) may all be chosen of degree \(n\) if only the series \[\sum_{\nu=1}^{\infty} \frac{|\alpha_{\nu}|}{|z_{\nu}|^{n+1}}\] converges. This case is the most important one for the applications. The meromorphic functions that arise in practice usually have simple poles.

Although the Mittag-Leffler Theorem can be used easily to expand a function with simple poles into partial fractions we would still have the problem of determining the entire function \(\omega(z)\) . In this special case we have a more direct method.

Let \(C\) be any simple closed curve which avoids the poles of \(f(z)\) . If \(z\) is any regular point in the interior of \(C\) then we have \[f(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} \,d \zeta-\sum \operatorname{Res} \frac{f(z_{\nu})}{z_{\nu}-z}\] where \(\operatorname{Res} \frac{f(z_{\nu})}{z_{\nu}-z}\) denotes the residue of \(\frac{f(\zeta)}{\zeta-z}\) at \(\zeta=z_{\nu}\) , and where the sum is taken over all the singularities \(z_{\nu}\) of \(f(z)\) in the interior of \(C\) . Now, since \(f(z)\) is assumed to have only simple poles, we have (by (1.24) ) \(\operatorname{Res} \frac{f(z_{\nu})}{z_{\nu}-z}=\frac{\operatorname{Res} f(z_{\nu})}{z_{\nu}-z}\) and thus we obtain the representation \[f(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} \,d z+\sum \frac{\operatorname{Res}f(z_{\nu})}{z-z_{\nu}}\] 

Since the poles are isolated we can surely find a sequence of closed curves \(C_{n}\) , each containing all the preceding ones in its interior, each avoiding all the poles of \(f(z)\) , such that the distance of \(C_{n}\) from the origin tends to infinity with \(n\) . If for some such sequence \(\lim _{n \rightarrow \infty} \int_{C_{n}} \frac{f(\zeta)}{\zeta-z} \,d \zeta=0\) , then denoting the poles in the ring-shaped domain between \(C_{n-1}\) and \(C_{n}\) by \(z_{\nu}^{(n)}\) we see that the series \[\tag{2.35} f(z)=\sum_{n=1}^{\infty}\left(\sum\frac{\operatorname{Res} f\big(z_{\nu}^{(n)}\big)}{z-z_{\nu}^{(n)}}\right)\] converges and gives the decomposition of the function into partial fractions.

Consider the special case where \(f(z)\) is the quotient of two polynomials \(f(z)=P(z) / Q(z)\) where the degree \(n\) of \(Q(z)\) is greater than that of \(P(z)\) and \(Q(z)\) has only simple roots \({z}_{1}, {z}_{2}, \ldots, {z}_{{n}}\) which are not roots for \({P}({z})\) . The above representation gives \begin{align} f(z)&=\frac{P(z)}{Q(z)}\\ &=\frac{1}{2 \pi{i}} \int_{C_{\rho}} \frac{P(\zeta) \,d \zeta}{Q(\zeta)(\zeta-z)}+\sum_{\nu=1}^{n} \frac{\operatorname{Res} f(z_{\nu})}{z-z_{\nu}} \end{align} where \(C_{\rho}\) is a circle containing all the \(z_{\nu}\) . Now by (1.24) \(\operatorname{Res} f(z_{\nu})=P(z_{\nu}) / Q'(z_{\nu})\) and since \(f(\zeta)\) tends to zero (cf. III, (3.23) ) as the radius of \(C_{\nu}\) tends to infinity, we obtain \[\tag{2.36} \frac{P(z)}{Q(z)}=\sum_{\nu=1}^{n} \frac{1}{z-z_{\nu}} \frac{P(z_{\nu})}{Q'(z_{\nu})},\] the familiar formula of Lagrange for the decomposition of a rational function into partial fractions.

The decomposition of a transcendental meromorphic function with only simple poles is well illustrated by the function \[f(z)=\pi\, \frac{\cos \pi z}{\sin \pi z}=\pi \cot \pi z.\] The poles of this function occur at the zeros of \(\sin \pi z\) . The function \(\sin \pi z\) has a simple zero at the origin, hence, in view of its periodicity, at all integral values \(z=0, \pm 1, \pm 2, \ldots\) . It has no other zeros. The cosine function has zeros at none of these points. Consequently \(f(z)\) has its poles at the integer points and these are all simple. We now show that \[\pi \cot \pi z=\sum_{\nu=-\infty}^{+\infty} \frac{1}{z-\nu}\] where by the notation \(\sum_{\nu=-\infty}^{+\infty}\) we mean \(\lim _{n \rightarrow \infty} \sum_{\nu=-n}^{+n}\) . For paths of integration we take squares \({S}_{{n}}\) with sides parallel to the coordinate axes and side lengths twice \(\lambda_{n}={n}+\frac{1}{2}\) . These clearly satisfy the necessary requirements. Since the function is periodic, the residues at all the poles are the same as at the origin and, plainly, \(\operatorname{Res} f(0)=1\) . Hence, in any square \(S_{n}\) \[f(z)=\pi \cot \pi z=\frac{1}{2 \pi i} \int_{S_{n}} \frac{\pi \cot \pi \zeta}{\zeta-z} \,d \zeta+\sum_{\nu = -n}^{+n} \frac{1}{z-\nu}.\] In order to complete the proof we need only show that the integral in this equation tends to zero as \(n \rightarrow \infty\) . Accordingly, denote the sides of \({S}_{n}\) by \({H}_{n}^{+}\) , \({H}_{n}^{-}\) , \({V}_{n}^{+}\) , and \({V}_{n}^{-}\) as in the diagram.

Now \(\pi \cot \pi \zeta\) is an odd function, that is \[\pi \cot \pi(-\zeta)=-\pi \cot \pi \zeta.\] Consequently \[2 \pi i I_{H}=\int_{H^{+}}+\int_{H^{-}}=\int_{H^{+}} \pi \cot \pi \zeta\Big(\frac{2 z}{\zeta^{2}-z^{2}}\Big) \,d \zeta\] and similarly \[2 \pi i I_{V}=\int_{V^{+}}+\int_{V^{-}}=\int_{V^{+}} \pi \cot \pi \zeta\Big(\frac{2 z}{\zeta^{2}-z^{2}}\Big) \,d \zeta.\] On \(H_{n}^{+}\) we have \(\zeta=x+i \lambda_{n}\) , \(-\lambda_{n} \leq x \leq \lambda_{n}\) . On \(V_{n}^{+}\) , \(\zeta=\lambda_{n}+i y\) , \(-\lambda_n \leq y \leq \lambda_n\) . Since \(\pi \cot \pi \zeta=\pi i\, \frac{e^{2 \pi i \zeta}+1}{e^{2 \pi i \zeta}-1}\) we have on \(H_{n}^{+}\) \[|\pi \cot\pi\zeta| = \left|\pi i\, \frac{e^{2\pi ix}e^{-2\pi(n + 1/2)} + 1}{e^{2\pi ix}e^{-2\pi(n + 1/2)} - 1}\right| < 2\pi.\] For \({n}\) sufficiently large, and on \({V}_{{n}}^{+}\) \[|\pi \cot \pi \zeta|=\left|\pi i\, \frac{-e^{-2 \pi y}+1}{-e^{-2 \pi y}-1}\right|<2 \pi\] since \(-e^{-2 \pi y}\) can only take on negative real values. Thus \(\pi \cot \pi \zeta\) is uniformly bounded on the \(S_{n}\) .

This provides us with the estimate for \(|I_{H}|\) \[|I_H| \leq 2|z|\left|\int_{H^+} \frac{d\zeta}{\zeta^2 - z^2}\right|\] and also for \(|I_{V}|\) . But \(|\zeta \pm z| \geq|\zeta|-|z| \geq \lambda_{n}-|z|\) , whence \[|I_{H}| \leq 2|z|\, \frac{2 \lambda_{n}}{\left(\lambda_{n}-|z|\right)^{2}}\] and this clearly goes to zero as \(\lambda_{n} \rightarrow \infty\) . The proof is now complete. We have shown that \[\tag{2.37} \pi \cot \pi z=\sum_{\nu = -\infty}^{+\infty} \frac{1}{z - \nu}.\] The periodicity of the function is exhibited by the fact that we may substitute \(\nu+1\) for \(\nu\) without changing the expansion. It is often convenient to use the formula in the form \[\tag{2.37a}\pi \cot \pi z=\frac{1}{z}+\sum_{n=1}^{\infty} \frac{2 z}{z^{2}-n^{2}}.\] 

Now that we have obtained a partial fractional representation for meromorphic functions we may hope to use it to prove something for the zeros of an entire function, for the zeros of \(f(z)\) may be interpreted as simple poles for \(f'(z) / f(z)\) . In fact we prove the Weierstrass Product Theorem .

The same method used in the proof of the theorem can be applied to obtain the product representation of \(\sin \pi z\) from the expansion of \(\pi \cot \pi z\) , (2.37) , since \(\pi \cot \pi z=\frac{d}{d z} \log \sin \pi z\) . From (2.37) we derive \begin{align} \frac{d}{d z}(\log \sin \pi z)-\frac{1}{z} & =\frac{d}{d z} \log \Big(\frac{\sin \pi z}{z}\Big) \\ & =\sum_{\nu=1}^{\infty} \frac{2 z}{z^{2}-\nu^{2}}. \end{align} In any finite region the tail end of this series converges uniformly. Hence we may integrate term-by-term. Taking the integral from \(0\) to \(z\) we obtain \[\left.\log \frac{\sin \pi z}{z}\right|_{0} ^{z}=\sum_{\nu=1}^{\infty}\Big(\log (z^{2}-\nu^{2})\Big|_{0}^{z}\Big).\] Now at \(z=0\) , \(\frac{\sin \pi z}{z}\) has a removable singularity so we complete it to a regular function. Thus \[\log \frac{\sin \pi z}{\pi z}=\sum_{\nu=1}^{\infty} \log \Big(1-\frac{z^{2}}{\nu^{2}}\Big).\] From this we obtain the convergent infinite product \[\tag{2.39} \sin \pi z=\pi z \prod_{\nu=1}^{\infty}\Big(1-\frac{z^{2}}{\nu^{2}}\Big),\] a formula discovered by Euler. If we set \(z=\frac{1}{2}\) , we obtain the well-known Wallis product: \begin{align} \frac{\pi}{2}&=\prod_{\nu=1}^{\infty} \frac{2\nu.2\nu}{(2 \nu-1).(2 \nu+1)}\\ &=\frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdots. \end{align} 

Exercises