Applications of Contour Integration and the Residue Theorem

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Thomas Wu
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Abstract

Complex Analysis has been a long-established field of mathematics, originating with Augustin-Louis Cauchy’s Cours’ D’analyse. Two of the most fundamental concepts in the field are Cauchy’s Theorem and the Residue Theorem. In this paper, we will investigate the applications of both theorems. More specifically, we will investigate the use of both theorems in evaluating integrals, deriving identities, and proving more complicated theorems, which, in turn, are used later on to prove various other results that are demonstrated in the paper.

Introduction

Complex Analysis is the field of mathematics that studies functions with complex domains and codomains. The field started to come about through Augustin-Louis Cauchy, whom published the first works regarding the extension of the concepts of real analysis, which include continuity, series, convergence, and integration, to the complex plane. Integration over the complex plane is perhaps the most notable aspect of complex analysis, as unlike standard integration, which is simply integrating a function over the real number line, contour integration requires integrating a function over a curve, such as a circle. Two of the most fundamental concepts developed by Cauchy regarding contour integration are Cauchy’s Theorem and the Residue Theorem.
Cauchy’s Theorem states the following: if f(z) is holomorphic in a region \Omega and \gamma is a closed curve in \Omega, then

    \[\int_\gamma f(z) \, dz = 0\]

According to1, Cauchy conjured and proved the first version of this theorem in 1822 for closed rectangular contours, and later expanded this claim to more general paths in 1825 (although Cauchy’s paper regarding the expansion of the theorem was not published until 1874). In 1846, Cauchy proved another version of this theorem using exact differentials and Green’s Theorem. In 1851, Bernhard Riemann avoided the use of exact differentials using Cauchy-Riemann Equations, but still employed Green’s Theorem to prove Cauchy’s Theorem for an analytic function with a continuous derivative. In 1900, Edouard Goursat expanded Cauchy’s Theorem to all analytic functions with existent derivatives. Since then, Cauchy’s Theorem has been used to prove many remarkable results in the field of complex analysis and serves as one of the most fundamental ideas in complex analysis.

One remarkable result that came about as a result of Cauchy’s Theorem is the Residue Theorem, which states that for a function f(z) holomorphic inside a region \gamma except for poles at z_{1},z_{2},z_{3}, \cdots,z_{n}, then:

    \[\int_\gamma f(z) \, dz = 2\pi i \sum_{k=1}^n \res_{z_k} f(z)\]

Both Cauchy’s Theorem and the Residue Theorem have vast applications to various other fields, such as Quantum Mechanics and Analytic Number Theory. As such, results that come about regarding Contour Integration, such as Cauchy’s Theorem and the Residue Theorem, are important for being able to expand upon knowledge involving fields that make heavy use of integrals, such as the two fields aforementioned above. In this paper, we will use both Cauchy’s Theorem and the Residue Theorem to demonstrate the vast applications of both theorems. These applications include the direct evaluation of various integrals, an extrapolation on an integral identity, and derivations of theorems and identities.

Preliminaries

It is important to note a few conventions used throughout the paper. First, call z_0 a pole of f(z) if \frac{1}{f(z)}=0 at z_0. Furthermore, call z_0 a pole of order n of f(z) if \left(z-z_0\right)^n f(z) doesn’t have a pole at z_0, but \left(z-z_0\right)^{n-1} f(z) does have a pole z_0. Next, we denote the residue of f(z) at z_0 as \operatorname{Res}_{z_0} f(z), and we evaluate it as:

    \[\operatorname{Res}{z_0} f(z)=\frac{1}{(n-1)!} \lim {z \rightarrow z_0} \frac{d^{n-1}}{d z^{n-1}}\left(\left(z-z_0\right)^n f(z)\right)\]

where n is the order of the pole at z_0. For example, the function f(z)=\frac{\left(1+z+z^2+z^3\right)(1+z)}{z^4} has a pole of order 4 at z=0, and the residue of f(z) at z=0 is calculated as

    \[\res_0 f(z)=\frac{1}{3!} \lim _{z \rightarrow 0} \frac{d^3}{dz^3}\left(z^4\left(\frac{\left(1+z+z^2+z^3\right)(1+z)}{z^4}\right)\right)=2\]

Finally, we define \operatorname{Arg}(z) as the principle argument of z; that is, we define \operatorname{Arg}(z) to be the angle t \in(-\pi, \pi] such that \frac{z}{|z|}=e^{i t}

Introductory Integrals

One of the classical applications of contour integration and the residue theorem is for evaluating definite integrals. One example of this is the following integral:

    \[\int_{-\infty}^\infty \frac{\cos(x)}{x^2 + 1} \, dx\]

which evaluates to \frac{\pi}{e}. To show this, we consider the function \frac{e^{i z}}{z^2+1}, as well as the idea that for real z, the real component of \frac{e^{i z}}{z^2+1} is simply \frac{\cos (x)}{x^2+1}. Next, we consider a semi-circular contour, as the integral over the diameter is essentially integrating f(z) over the real axis, and |f(z)|=\frac{1}{1+z^2} will behave similarly to \frac{1}{r^2}, meaning that the integral over the circumference will be near negligible as r approaches infinity. All of the above facts motivate us to integrate the function f(z)=\frac{e^{i z}}{z^2+1} over a semi-circular contour with large radius r, as shown in Figure 1.

                                             Figure 1: Semicircular Contour of Radius

Note that for all values of r>1, the only pole of f(z) is at z=i (which is a first-order pole). Using the residue theorem, we get that:

(1)   \begin{equation*} \int_\gamma f(z) \, dz + \int_C f(z) \, dz = 2\pi i \res_i f(z)\end{equation*}

Taking the integral of f(z) over \gamma, we substitute z=x \Rightarrow \mathrm{~d} z=\mathrm{d} x, which yields that the integral of f(z) over \gamma equals \int_{-r}^r f(x) \mathrm{d} x

We now consider f(z)=\frac{e^{i z}}{z^2+1} over C. On C, z=r e^{i \theta} \Rightarrow \mathrm{~d} z= ir e^{i \theta} \mathrm{~d} \theta for \theta going from 0 to \pi. Plugging this in, we have that \int_C f(z) \mathrm{d} z=\int_0^\pi \frac{\text { ire }^{i r e^\theta+i \theta}}{r^2 e^{2 i \theta}+1} \mathrm{~d} \theta. Considering the modulus of the integrand over the semi-circular arc of radius r, we have that \left|\frac{i r e^{i r e^\theta+i \theta}}{r^2 e^{2 i \theta}+1}\right| \leq \frac{r}{r^2-1}. We see that the modulus of the integrand approaches 0 as r goes to infinity. This implies that \int_C f(z) \mathrm{d} z=0. We have that: \operatorname{Res}<em>i f(z)=\lim </em>{z \rightarrow i}(z-i) \frac{e^{i z}}{z^2+1}=\lim _{z \rightarrow i} \frac{e^{i z}}{z+i}=-\frac{i}{2 e}. This means that, letting the radius of the semicircular contour tend to infinity, (1) ultimately becomes:

    \[\int_{-\infty}^\infty \frac{e^{ix}}{x^2 + 1} \, dx = \frac{\pi}{e}\]

Since x is being integrated along the real axis, taking the real component of both sides yields that:

    \[\int_{-\infty}^\infty \frac{\cos(x)}{x^2 + 1} \, dx = \frac{\pi}{e}\]

Gaussian Integral

Introduction and Derivation of Complex Gaussian
The well-known Gaussian integral identity states that:

    \[\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}\]

Now, consider offsetting exponent of the initial integrand by a positive real constant a^2, leading to the integrand becoming e^{-a^2 x^2}. By letting u=a x, one can find that

(2)   \begin{equation*}\int_{-\infty}^{\infty} e^{-a^2 x^2} \mathrm{~d} x=\frac{\sqrt{\pi}}{a}\end{equation*}

However, this begs the following question: does this equation hold when a is an imaginary number?
Inspired by2, we first consider the function: f(z)=e^{-z^2} and consider a contour consisting of a circular arc with angle \theta, as shown in Figure 2

                        Figure 2: Circular Arc Contour of radius  and angle

One reason for choosing this function and contour is that there are no poles in the contour, so, by Cauchy’s Theorem, we have that the integral of f(z) over the entire contour equals 0. We also have 3 segments for this integral, which are all convenient in our evaluation of the general integral (1). When we expand our Cauchy Theorem expression based on segments, we have that:

(3)   \begin{equation*} \int_{\gamma_1} f(z) \, dz + \int_C f(z) \, dz + \int_{\gamma_2} f(z) \, dz = 0 \end{equation*}

Over \gamma_1, it is equivalent to integrating the function over the real axis from x=0 to x=r, so we have that \int_0^r e^{-x^2} \mathrm{~d} x.

Over C, we have that the angle varies, meaning that z=r e^{i \phi}, where \phi varies between 0 and \theta. Now, notice that for |\operatorname{Arg}(z)|>\frac{\pi}{4}, \mathfrak{R}\left(-z^2\right)>0, which, considering the fact that z^2 is directly proportional to r^2, implies that e^{-z^2} approaches infinity if |\operatorname{Arg}(z)|>\frac{\pi}{4}. To avoid complications regarding the integration of our curve over C, we must restrict \theta so that |\operatorname{Arg}(z)| is always less than or equal to \frac{\pi}{4}. Therefore, we impose a further restriction that \theta \in\left[0, \frac{\pi}{4}\right].

Over \gamma_2, z is represented by te^{i\theta}, where t varies from r to 0. This means that:

    \[\int_{\gamma_2} f(z) \, dz = e^{i\theta} \int_r^0 e^{-t^2 e^{2i\theta}} \, dt\]

Letting r tend to infinity, (2) ultimately becomes:

    \[\int_0^\infty e^{-x^2} \, dx + e^{i\theta} \int_\infty^0 e^{-t^2 e^{2i\theta}} \, dt = 0 \quad \Rightarrow \quad \int_0^\infty e^{-t^2 e^{2i\theta}} \, dt = \frac{\sqrt{\pi}}{2} e^{-i\theta}\]

For \theta \in \left[0, \frac{\pi}{4}\right], the derived expression is similar to (1), and shows that (1) can be extrapolated for complex numbers a that satisfy \text{Arg}(a) \in \left[0, \frac{\pi}{4}\right]. Furthermore, we can extrapolate this and show that (1) works for all complex numbers a with |\text{Arg}(a)| \leq \frac{\pi}{4} by considering the following contour demonstrated in Figure 3.

Figure 3: Circular Arc Contour of Radius r and Angle - \theta

Gaussian with Complex Center

Now, we aim to expand upon the formula derived in the previous subsection, and prove that:

    \[\int_{-\infty}^\infty e^{-a^2 (t + \omega)^2} \, dt = \int_{-\infty}^\infty e^{-a^2 t^2} \, dt = \frac{\sqrt{\pi}}{a}\]

For complex constants a and \omega with |\text{Arg}(a)| \leq \frac{\pi}{4}, we proceed as follows. To do this, first assume, without loss of generality, that \omega is a purely imaginary number (i.e., \omega = ki for real number k). This is due to the fact that if \omega does have a nonzero real portion, we write \omega = x + yi for x, y \in \mathbb{R} and let u = t + x, which yields an integral with a purely imaginary center.

Now, we consider the function f(z) = e^{-a^2 z^2}, where a is a complex number such that |\text{Arg}(a)| \leq \frac{\pi}{4}, and a rectangular contour with length 2r and width \omega, as shown in Figure 4.

                     Figure 4: Rectangular Contour of Length 2r and Width \omega

There are a few reasons for choosing this contour. First, f(z) has no poles in the contour for all values of r and all purely imaginary values of \omega. Secondly, since we stipulated that |\text{Arg}(a)| \leq \frac{\pi}{4}, the integrals over the widths of the rectangle are going to be negligible since e^{-a^2 z^2} will approach 0 as r approaches infinity. Finally, the integrals over the lengths will provide a convenient method for evaluating the desired integral.

Since f(z) has no poles inside the contour, by Cauchy’s Theorem, we have:

(4)   \begin{equation*}\int_{\gamma_1} f(z) \, dz + \int_{\gamma_2} f(z) \, dz + \int_{\gamma_3} f(z) \, dz + \int_{\gamma_4} f(z) \, dz = 0\end{equation*}

Over \gamma_{1}, we have that z=x \implies dz=dx:

    \[\int_{\gamma_1} f(z) \, dz = \int_{-r}^{r} e^{-a^2 x^2} \, dx\]

Over \gamma_2, z = r + iy \quad \Rightarrow \quad dz = i \, dy, where y ranges from 0 to \omega. Plugging z into f(z) yields e^{-a^2 (r + iy)^2}, meaning that our integral over \gamma_2 is \int_0^\omega i e^{-a^2 (r + iy)^2} \, dy. Now, for |\text{Arg}(a)| \leq \frac{\pi}{4}, i e^{-a^2 (r + iy)^2} will approach 0 as r approaches infinity, which indicates that \int_{\gamma_2} f(z) \, dz = 0 as r approaches infinity.

Over \gamma_3, we have that z = x + \omega \quad \Rightarrow \quad dz = dx, where x ranges from r to -r. Therefore, our integral over \gamma_3 becomes:

    \[\int_{\gamma_3} f(z) \, dz = \int_r^{-r} e^{-a^2 (x + \omega)^2} \, dx\]

Over \gamma_4, z = -r + iy \quad \Rightarrow \quad dz = i \, dy, where y ranges from \omega to 0. Plugging z into f(z) yields e^{-a^2 (-r + iy)^2}, meaning that our integral over \gamma_4 is \int_\omega^0 i e^{-a^2 (-r + iy)^2} \, dy. Now, for |\text{Arg}(a)| \leq \frac{\pi}{4}, i e^{-a^2 (-r + iy)^2} will approach 0 as r approaches infinity, which indicates that \int_{\gamma_4} f(z) \, dz = 0 as r approaches infinity.

Letting r tend to infinity, we have that (1) becomes:

    \[\int_{-\infty}^\infty e^{-a^2 x^2} \, dx - \int_{-\infty}^\infty e^{-a^2 (x + \omega)^2} \, dx = 0 \quad \Rightarrow\]

    \[\quad \int_{-\infty}^\infty e^{-a^2 x^2} \, dx = \int_{-\infty}^\infty e^{-a^2 (x + \omega)^2} \, dx\]

which was the desired equality.

Application of the Complex Gaussian

Typically, the standard Gaussian Integral has its applications in the field of Statistics, with variations of it being related to normal distributions. The integral representing a normal distribution with variance \sigma^2 and mean \mu is given as:

    \[\frac{1}{\sigma \sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{(t - \mu)^2}{2\sigma^2}} \, dt\]

The Complex Gaussian identities that we derived in earlier subsection have many applications to other fields, primarily quantum mechanics. One such example is with the following integral equality:

(5)   \begin{equation*}\int_{-\infty}^\infty e^{-\frac{1}{2} ax^2 + iJx} \, dx = \sqrt{\frac{2\pi}{a}} e^{-\frac{J^2}{2a}} \end{equation*}

For complex constants a and J where the real part of a is nonnegative. To show this, we try to get the integrand of our left-hand expression to be of the form derived in the previous section. We can do this by considering the exponent of the exponential in the integrand and manipulating it:

    \[-\frac{1}{2} ax^2 + iJx = -\frac{1}{2} a \left( x^2 - \frac{2iJ}{a} x \right) = -\frac{1}{2} a \left( x - \frac{iJ}{a} \right)^2 - \frac{J^2}{2a}.\]

Therefore, we can rewrite our integral as:

    \[e^{-\frac{J^2}{2a}} \int_{-\infty}^\infty e^{-\frac{a}{2} \left( x - \frac{iJ}{a} \right)^2} \, dx = \sqrt{\frac{2\pi}{a}} e^{-\frac{J^2}{2a}}\]

where the last equality references the result from the previous section.

This equation has many applications in the field of quantum mechanics. For one, if the Fourier Transform of e^{-\frac{a}{2} x^2} is f(\xi), then (1) is simply f\left(-\frac{J}{2\pi}\right). Additionally, (1) demonstrates the uncertainty principle, as J and x are conjugate variables. This means that as a increases, the narrower the Gaussian in x becomes (due to the decrease in the variance of the distribution of x) and the wider the Gaussian in J becomes (as the variance of the distribution of J increases as a increases).

Another application of our Gaussian Integral results is with respect to the propagator of a one dimensional free particle. The probabiliity that a particle will travel from x to x' in a time t is given by:

(6)   \begin{equation*}K(x, x'; t) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ik(x - x') - \frac{ihk^2 t}{2m}} \, dk\end{equation*}

where h is Planck’s constant and m is the mass of the particle. We can factor the exponent in an attempt to apply the result from 4.2:

    \[-\frac{ihk^2 t}{2m} + ik(x - x') = -i \left(k \sqrt{\frac{ht}{2m}} - \frac{x - x'}{2 \sqrt{\frac{ht}{2m}}}\right)^2 + i \frac{m(x - x')^2}{2ht}.\]

Thus, we can rewrite the given equation as:

    \[\frac{1}{2\pi} \int_{-\infty}^\infty e^{ -i \left( k \sqrt{\frac{ht}{2m}} - \frac{x - x'}{2 \sqrt{\frac{ht}{2m}}} \right)^2 + i \frac{m(x - x')^2}{2ht} } \,\]

    \[dk= \frac{e^{i \frac{m(x - x')^2}{2ht}}}{2\pi} \int_{-\infty}^\infty e^{-i \left( k \sqrt{\frac{ht}{2m}} - \frac{x - x'}{2 \sqrt{\frac{ht}{2m}}} \right)^2} \, dk\]

Using the result in 4.2 with a = e^{i \frac{\pi}{4}} \sqrt{\frac{ht}{2m}}, and \omega = \frac{m(x - x')}{ht}, we find that we can evaluate the given integral to be \sqrt{\frac{m}{2\pi i ht}} e^{i \frac{m(x - x')^2}{2ht}}, providing a function that determines the probability that a particle of mass m will travel from x to x' in time t.

Derivation of the Euler Reflection Formula

The Euler Reflection Formula states that, for some complex number x that is not an integer,

    \[\Gamma(x) \Gamma(1 - x) = \pi \csc(\pi x),\]


where \Gamma(x) is the well-known Gamma function. While many proofs for the theorem use Weierstrass products and the Weierstrass form of the Gamma function, the theorem can also be shown by using the Residue Theorem and the Beta Function.

The Beta Function is given as

    \[\beta(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)}.\]


Solving a double integral through change of variables, one can obtain that

    \[\beta(x, y) = \int_0^1 t^{x-1} (1 - t)^{y-1} \, dt.\]


From there, letting t = \frac{v}{1 - v} yields the following expression:

    \[\beta(x, y) = \int_0^\infty \frac{v^{x-1}}{(1 + v)^{x + y}} \, dv\]

Now, we let v = e^u \quad \Rightarrow \quad dv = e^u \, du. This makes our integral into

    \[\beta(x, y) = \int_{-\infty}^{\infty} \frac{e^{xu}}{(1 + e^u)^{x + y}} \, du\]


Now, let x = a, where a is a real number such that 0 < a < 1, and y = 1 - a, which turns the integral into

    \[\beta(a, 1-a) = \int_{-\infty}^{\infty} \frac{e^{au}}{1 + e^u} \, du.\]

From here, we proceed with this integral the same way as described in2. Consider the function f(z) = \frac{e^{az}}{1 + e^z}, as well as the rectangular contour of length 2r (where r is very large) and width 2\pi, as shown in Figure 5:

Figure 5: Rectangular Contour of Length 2r and Width 2 \pi i}

Notice that, for all r, f(z) only has a first-order pole at z = \pi i. This provides us motivation for using a rectangular contour, as well as the fact that the function can be rewritten as \frac{1}{e^{(1-a)z} + e^{-az}}, which means that the integral over the widths of the rectangle will be nearly negligible as r approaches infinity. Furthermore, the lengths of the rectangular contour allow us a method for solving the given integral, which further justifies the use of a rectangular contour. Now, by the Residue theorem, we have that \text{Res}_{\pi i} f(z) equals:

(7)   \begin{equation*}\int_{\gamma_1} f(z) \, dz + \int_{\gamma_2} f(z) \, dz + \int_{\gamma_3} f(z) \, dz + \int_{\gamma_4} f(z) \, dz\end{equation*}

The integral over \gamma_1 is equivalent to integrating f(z) from x=-r to x=r, so it can be represented as:

    \[\int_{-r}^{r} \frac{e^{ax}}{1+e^x} \, dx\]

Over both \gamma_2 and \gamma_4, the imaginary components of z vary, and the real components are constant, being r and -r, respectively. We can rewrite z in these situations as r+yi and -r+yi, respectively. In both situations, dz=i \, dy. We first rewrite f(z) as \frac{1}{e^{(1-a)z} + e^{-az}} and note that, because a \in (0,1), a and 1-a have positive real portions. So, our integrals over \gamma_2 and \gamma_4 are:

    \[\int_0^{2\pi} i f(r+yi) \, dy \quad \text{and} \quad \int_0^{2\pi} i f(-r+yi) \, dy,\]


respectively.

Now, note that |f(z)| = \left|\frac{1}{e^{-az} + e^{(1-a)z}}\right| = \frac{1}{|e^{-az} + e^{(1-a)z}|}. Over \gamma_2, z = r + yi, meaning that e^{(1-a)z} will tend to infinity as r approaches infinity, and e^{-az} will tend to 0 as r approaches infinity, so |i f(z)| = |f(z)| over \gamma_2 will tend to 0 as r approaches infinity. Similarly, over \gamma_4, z = -r + yi, meaning that e^{-az} will tend to infinity as r approaches infinity, and e^{(1-a)z} will tend to 0 as r approaches infinity, so |i f(z)| = |f(z)| over \gamma_4 will tend to 0 as r approaches infinity. Both of these statements lead to the conclusions that:

    \[\int_{\gamma_2} f(z) \, dz = 0 \quad \text{and} \quad \int_{\gamma_4} f(z) \, dz = 0,\]


as r approaches infinity.

Over \gamma_3, z = x + 2\pi i, where x goes from r to -r. Substituting this into our integral yields that, over \gamma_3:

    \[\int_{\gamma_3} f(z) \, dz = \int_r^{-r} \frac{e^{a(x+2\pi i)}}{1+e^x} \, dx = -e^{2\pi ai} \int_{-r}^{r} \frac{e^{ax}}{1+e^x} \, dx.\]

We can evaluate \text{Res}<em>{\pi i} f(z) as:

    \[\lim</em>{z \to \pi i} \frac{(z-\pi i)}{(e^z - e^{\pi i})(e^{az})} = -e^{a\pi i}.\]

Using our previous calculations, we can see that as r approaches infinity, (1) will ultimately come out to:

    \[\int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^x} \, dx - e^{2\pi ai} \int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^x} \, dx = -2\pi i e^{a\pi i},\]


which yields the conclusion that:

    \[\int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^x} \, dx = \frac{-2\pi i e^{a\pi i}}{1 - e^{2\pi ai}} = \frac{2\pi i}{e^{a\pi i} - e^{-a\pi i}} = \pi \csc(\pi a).\]

Since our original integral was shown, by the Beta function, to be equal to \frac{\Gamma(a)\Gamma(1-a)}{\Gamma(1)}, and \Gamma(1) = 1, this proves that:

    \[\Gamma(a)\Gamma(1-a) = \pi \csc(\pi a) \quad \text{for all} \quad a \in (0,1).\]

Furthermore, both \Gamma(x)\Gamma(1-x) and \pi \csc(\pi x) have the same set of singular points (where x is an integer). Therefore, on the complement of the integers, both are holomorphic. Moreover, they are equal on the open set defined by R(x) \in (0,1). So these are two holomorphic functions which are equal on a nonempty open set, hence they are equal on the complement of integers. In other words, we have that:

\Gamma(x)\Gamma(1-x) = \pi \csc(\pi x) for all non-integral complex numbers x, as desired.

Jensen’s Formula

Introduction and Derivation of Jensen’s Formula

The reference2 mentions an interesting claim: Let \Omega be an open set that contains the closure of a disc D_R centered at the origin. Let a function f be defined such that it is holomorphic in \Omega, f(0) \neq 0, and it vanishes nowhere on the circle C_R centered at the origin. If z_1, z_2, z_3, \dots, z_N are the zeros of f counted with multiplicities (meaning that the zeros are not necessarily distinct) inside D_R, then:

    \[\log|f(0)| = \sum_{k=1}^N \log\left(\frac{|z_k|}{R}\right) + \frac{1}{2\pi} \int_0^{2\pi} \log|f(Re^{i\theta})| \, d\theta.\]

This claim is known as Jensen’s formula, and the heart of the proof lies in the concepts of Cauchy’s Theorem and the Residue Theorem.

To prove Jensen’s formula, we first start with the following claim: If f is holomorphic in an open set that contains the closure of a disc D, and C denotes the boundary circle of this disc with positive orientation, then:

(8)   \begin{equation*}f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_0} dz \quad \end{equation*}


This result is immediate due to the Residue Theorem, as 2\pi i \, \text{Res}<em>{z_0} f(z) = \int_C \frac{f(z)}{z - z_0} since the only pole of \frac{f(z)}{z - z_0} is at z = z_0 (which is a first-order pole), and we can find that \text{Res}</em>{z_0} f(z) = \lim_{z \to z_0} \frac{(z - z_0)f(z)}{z - z_0} = f(z_0).

Next, we will show that, for a function f holomorphic in a disc D_R and 0 < r < R:

(9)   \begin{equation*}f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) \, d\theta\end{equation*}

To show this, we start with (1): f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z - z_0} dz, where C is a boundary circle of radius r \in (0, R) centered at z_0. Then, consider the fact that every point on C can be represented as z_0 + re^{i\theta}, where \theta is between 0 and 2\pi. If we let z = z_0 + re^{i\theta}, then dz = ire^{i\theta} d\theta. Our equation for f(z_0) becomes:

    \[f(z_0) = \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z_0 + re^{i\theta})}{re^{i\theta}} ire^{i\theta} d\theta = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) \, d\theta,\]

as desired.

Now, consider the function \log |g(z)|, where g(z) is holomorphic in \Omega and vanishes nowhere inside \Omega (that is, g(z) has no zeroes in \Omega). It is given that \log |g(z)| is also holomorphic in \Omega. Using (2), and plugging in z_0 = 0, we have that:

(10)   \begin{equation*}\log |g(0)| = \frac{1}{2\pi} \int_0^{2\pi} \log |g(re^{i\theta})| \, d\theta\end{equation*}

Next, consider the function h(z) = z - \omega, where \omega lies in D_R. We will show that:

(11)   \begin{equation*}\log |h(0)| = \log \left( \frac{|\omega|}{R} \right) + \frac{1}{2\pi} \int_0^{2\pi} \log |h(Re^{i\theta})| \, d\theta\end{equation*}

To do this, first note that \log |h(0)| = \log |\omega| and that \log |h(Re^{i\theta})| = \log |Re^{i\theta} - \omega|. Now, note that:

    \[\frac{1}{2\pi} \int_0^{2\pi} \log |Re^{i\theta} - \omega| \, d\theta = \log (R) + \frac{1}{2\pi} \int_0^{2\pi} \log |e^{i\theta} - \omega/R| \, d\theta.\]

From here, we will show that

    \[\int_0^{2\pi} \log |e^{i\theta} - a| \, d\theta = 0 \quad \text{if} \quad |a| < 1.\]

We first rewrite the integral, using a \theta \to -\theta substitution, as \int_0^{2\pi} \log |1 - ae^{i\theta}| \, d\theta. Since \log |1 - az| is holomorphic inside D for |a| < 1, we have, by plugging in the function \log |1 - az| and z_0 = 0 into (2), that:

    \[\frac{1}{2\pi} \int_0^{2\pi} \log |1 - ae^{i\theta}| \, d\theta = \log(1) = 0,\]

meaning that our desired claim that \int_0^{2\pi} \log |e^{i\theta} - a| \, d\theta = 0 if |a| < 1 holds. Since \log |h(0)| = \log |\omega| and

    \[\frac{1}{2\pi} \int_0^{2\pi} \log |Re^{i\theta} - \omega| \, d\theta = \log(R) + \frac{1}{2\pi} \int_0^{2\pi} \log |e^{i\theta} - \omega/R| \,\]

    \[d\theta = \log(R),\]

we have that (4) holds.

Now, note that both (3) and (4) satisfy Jensen’s Formula. We will now show that if two functions, f_1 and f_2, each satisfy Jensen’s formula, then the product of the two functions, f_1 f_2, will also satisfy Jensen’s formula. To see this, we first let the {a_1,a_2,a_3,\dots,a_n} be the zeros of f_1 and {b_1,b_2,b_3,\dots,b_m} be the zeros of f_2. Given that both f_1 and f_2 satisfy Jensen’s formula, we have that:

    \[\log |f_1 (0)| = \sum_{k=1}^n \log \left( \frac{|a_k|}{R} \right) + \frac{1}{2\pi} \int_0^{2\pi} \log |f_1(Re^{i\theta})| \, d\theta\]

    \[\log |f_2 (0)| = \sum_{d=1}^m \log \left( \frac{|b_d|}{R} \right) + \frac{1}{2\pi} \int_0^{2\pi} \log |f_2(Re^{i\theta})| \, d\theta\]

Now, adding both equations yields and using real logarithm properties (since both |f_1(z)| and |f_2(z)| are real numbers):

    \[\log |f_1(0) f_2(0)| = \sum_{k=1}^n \log \left( \frac{|a_k|}{R} \right) + \sum_{d=1}^m \log \left( \frac{|b_d|}{R} \right) +\]

    \[\frac{1}{2\pi} \int_0^{2\pi} \log |f_1(Re^{i\theta}) f_2(Re^{i\theta})| \, d\theta\]

Considering the fact that the union of the zeros of f_1 and the zeros of f_2 make up the zeros of f_1 f_2 (all of the counted zeroes are counted with multiplicity), we have shown that Jensen’s formula holds for f_1 f_2. This completes the proof for Jensen’s Theorem, as every holomorphic function in \Omega is analytic, and can therefore be written as a polynomial, meaning that a function f with zeros in \Omega at z_1, z_2, z_3, \dots, z_n can be written as f(z) = (z - z_1)(z - z_2)(z - z_3)\dots(z - z_n)g(z), where g(z) vanishes nowhere inside \Omega, and we can then use the last claim to show that all functions f(z) will satisfy Jensen’s formula.

Establishing a Bound for the Number of Zeros.

WWe can use Jensen’s Formula to establish a bound for the number of zeros that a function f has in a disc. First, without loss of generality, consider a disc inside a region \Omega, D_R, of radius R centered at the origin. Note that we can apply this to discs centered elsewhere due to translation (that is, we can declare a new function g(z) such that g(z) = f(z - z_0), where z_0 is any point in the complex plane that we desire to center our disc at). Furthermore, assume that f is holomorphic inside \Omega. We aim to establish a bound for the number of zeros inside a disc D_r concentric to D_R with radius r \in (0, R).

First, we assume that, in D_R, \lvert f(z) \rvert \leq M (in other words, f has a maximum modulus of M inside D_R). Using this bound, we can find that:

(12)   \begin{equation*}\frac{1}{2\pi} \int_0^{2\pi} \log \lvert f(Re^{i\theta}) \rvert \, d\theta \leq \frac{1}{2\pi} \int_0^{2\pi} \log(M) \, d\theta = \log(M)\end{equation*}

Next, let n be the number of zeros inside D_r and N be the number of zeros inside D_R. Then, consider \sum_{k=1}^N \log \left( \frac{\lvert z_k \rvert}{R} \right). For the n zeros that lie in D_r, \lvert z_k \rvert < r, so \log \left( \frac{\lvert z_k \rvert}{R} \right) < \log \left( \frac{r}{R} \right). For all other zeros (that is, the ones that lie outside D_r), we have that \log \left( \frac{\lvert z_k \rvert}{R} \right) < 0. Therefore, we create a new bound:

(13)   \begin{equation*}\sum_{k=1}^N \log \left( \frac{\lvert z_k \rvert}{R} \right) < \sum_{k=1}^n \log \left( \frac{r}{R} \right) = n \log \left( \frac{r}{R} \right)\end{equation*}

Adding (1) and (2), we get that:

    \[\log(M) + n \log \left( \frac{r}{R} \right) > \frac{1}{2\pi} \int_0^{2\pi} \log \lvert f(Re^{i\theta}) \rvert \, d\theta + \sum_{k=1}^N \log \left( \frac{\lvert z_k \rvert}{R} \right)\]

    \[= \log \lvert f(0) \rvert,\]


where the last equality follows from an application of Jensen’s formula. We can rearrange the inequality to get:

    \[n \log \left( \frac{r}{R} \right) > \log \left( \frac{\lvert f(0) \rvert}{M} \right).\]


Negating both sides and using properties of the logarithm gives:

    \[n \log \left( \frac{R}{r} \right) < \log \left( \frac{M}{\lvert f(0) \rvert} \right).\]

Since \log \left( \frac{R}{r} \right) > 0, dividing by \log \left( \frac{R}{r} \right) yields:

    \[n < \frac{\log \left( \frac{M}{\lvert f(0) \rvert} \right)}{\log \left( \frac{R}{r} \right)}.\]

Alternatively, this can be written as:

    \[#(f^{-1}({0}) \cap D_r) < \frac{\log \left( \frac{\max_{z \in D_R} \lvert f(z) \rvert }{\lvert f(0) \rvert }\right) }{\log \left( \frac{R}{r} \right)}.\]

The inequality above provides an upper bound on the number of roots that a function f, with the properties of being holomorphic inside \Omega (which contains D_R), has inside a disc D_r \subset D_R, as desired.

References

  1. Scott, E.A. (1974). Cauchy’s Integral Theorem: A Historical Development of Its Proof [↩]
  2. Stein, E., & Shakarchi, R. (2003). Complex Analysis. Princeton University Press. [↩] [↩] [↩]

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