The Residue Theorem

Section 8.1 The Residue Theorem

The Cauchy integral formulas given in Section 6.5 are useful in evaluating contour integrals over a simple closed contour $C$ where the integrand has the form $\frac{f(z)}{(z-z_0)^k}$ and $f$ is an analytic function. In this case, the singularity of the integrand is at worst a pole of order $k$ at $z_0\text{.}$ We begin this section by extending this result to integrals that have a finite number of isolated singularities inside the contour $C\text{.}$ This new method can be used in cases where the integrand has an essential singularity at $z_0$ and is an important extension of the previous method.

Definition 8.1.1. Residue.

Let $f$ have a nonremovable isolated singularity at the point $z_0\text{.}$ Then $f$ has the Laurent series representation for all $z$ in some disk $D_{R}^*(z_0)$ given by $f(z) = \sum\limits_{n=-\infty}^{\infty}a_n(z-z_0)^n\text{.}$ The coefficient $a_{-1}$ is called the residue of $f$ at $z_0$. We use the notation

\begin{equation*} \mathrm{Res}[f,z_0] = a_{-1}\text{.} \end{equation*}

Example 8.1.2.

If $f(z) = \exp(\frac{2}{z})\text{,}$ then the Laurent series of $f$ about the point 0 has the form

\begin{equation*} f(z) = \exp\left(\frac{2}{z}\right) = 1+\frac{2}{z}+\frac{2^2}{2!z^2} + \frac{2^3}{3!z^3}+\cdots\\text{,.} \end{equation*}

and $\mathrm{Res}[f,0] =a_{-1}=2\text{.}$

Example 8.1.3.

Find $\mathrm{Res}[g,0]$ if $g(z)=\frac{3}{2z+z^2-z^3}\text{.}$

Solution.

Using the techniques of Example 7.3.9, we find that $g$ has three Laurent series representations involving powers of $z\text{.}$ In the punctured disk $D_1^*(0)\text{,}$ $g(z) = \sum\limits_{n=0}^{\infty}[(-1)^n+\frac{1}{2^{n+1}}]z^{n-1}\text{.}$ Computing the first few coefficients, we obtain

\begin{equation*} g(z) = \frac{3}{2}\frac{1}{z}-\frac{3}{4}+\frac{9}{8}z-\frac{15}{16}z^2+\cdots\text{.} \end{equation*}

Therefore, $\mathrm{Res}[g,0] =a_{-1}=\frac{3}{2}\text{.}$

Recall that, for a function $f$ analytic in $D_{R}^*(z_0)$ and for any $r$ with $0\lt r\lt R\text{,}$ the Laurent series coefficients of $f$ are given by

\begin{equation} a_n = \frac{1}{2\pi i}\int\limits_{C_{r}^{+}(z_0)}\frac{f(\xi)}{(\xi -z_0)^{n+1}}\,d\xi \text{ for } n=0, \, \pm 1, \, \pm 2, \, \ldots\text{,}\tag{8.1.1} \end{equation}

where $C_{r}^{+}(z_0)$ denotes the circle $\{z:|z-z_0| =r\}$ with positive orientation. This result gives us an important fact concerning $\mathrm{Res}[f,z_0]\text{.}$ If we set $n=-1$ in Equation (8.1.1) and replace $C_{r}^{+}(z_0)$ with any positively oriented simple closed contour $C$ containing $z_0\text{,}$ provided $z_0$ is the still only singularity of $f$ that lies inside $C\text{,}$ then we obtain

\begin{equation} \int\limits_{C}f(\xi )\,d\xi = 2\pi ia_{-1}=2\pi i\mathrm{Res}[f,z_0]\text{.}\tag{8.1.2} \end{equation}

If we are able to find the Laurent series expansion for $f\text{,}$ then Equation (8.1.2) gives us an important tool for evaluating contour integrals.

Example 8.1.4.

Evaluate $\int\limits_{C_1^{+}(0)}\exp(\frac{2}{z})\,dz\text{.}$

Solution.

Example 8.1.2 showed that the residue of $f(z) = \exp\left(\frac{2}{z}\right)$ at $z_0=0$ is $\mathrm{Res}[f,0] =2\text{.}$ Using Equation (8.1.2), we get

\begin{equation*} \int\limits_{C_1^{+}(0)}\exp\left(\frac{2}{z}\right)dz = 2\pi i\mathrm{Res}[f,0] =4\pi i \end{equation*}

Theorem 8.1.5. Cauchy ’s residue theorem.

Let $D$ be a simply connected domain and let $C$ be a simple closed positively oriented contour that lies in $D\text{.}$ If $f$ is analytic inside $C$ and on $C\text{,}$ except at the points $z_1, \, z_2, \ldots , z_n$ that lie inside $C\text{,}$ then

\begin{equation*} \int\limits_{C}f(z)\,dz = 2\pi i\sum\limits_{k=1}^n\mathrm{Res}[f,z_k]\text{.} \end{equation*}

The situation is illustrated in Figure 8.1.6.

Figure 8.1.6. The domain $D$ and contour $C$ and the singular points $z_1, \ z_2, \ \ldots, \ z_n$ in the statement of Cauchy ’s residue theorem

Proof.

Since there are a finite number of singular points inside $C\text{,}$ there exists an $r>0$ such that the positively oriented circles $C_k=C_{r}^{+}(z_k)\text{,}$ for $k=1,2,\ldots,n\text{,}$ are mutually disjoint and all lie inside $C\text{.}$ From the extended Cauchy-Goursat theorem (Theorem 6.3.18 on page 6.3.18), it follows that

\begin{equation*} \int\limits_{C}f(z)\,dz = \sum\limits_{k=1}^n\int\limits_{C_k}f(z)\,dz\text{.} \end{equation*}

The function $f$ is analytic in a punctured disk with center $z_k$ that contains the circle $C_k\text{,}$ so we can use Equation (8.1.2) to obtain

\begin{equation*} \int\limits_{C_k}f(z)\,dz=2\pi i\mathrm{Res}[f,z_k], \text{ for } k=1,2,\ldots ,n\text{.} \end{equation*}

Combining the last two equations gives the desired result.

The domain $D$ and contour $C$ and the singular points $z_1, \, z_2, \, \ldots, \, z_n$ in the statement of Cauchy ’s residue theorem.

The calculation of a Laurent series expansion is tedious in most circumstances. As the residue at $z_0$ involves only the coefficient $a_{-1}$ in the Laurent expansion, we seek a method to calculate the residue from special information about the nature of the singularity at $z_0\text{.}$

If $f$ has a removable singularity at $z_0\text{,}$ then $a_{-n}=0\text{,}$ for $n=1,2,\ldots\text{.}$ Therefore, $\mathrm{Res}[f,z_0] =0\text{.}$ The following theorem gives methods for evaluating residues at poles.

Theorem 8.1.7. Residues at Poles.

i.

If $f$ has a simple pole at $z_0\text{,}$ then

\begin{equation*} \mathrm{Res}[f,z_0] =\lim\limits_{z \to z_0}(z-z_0)f(z)\text{.} \end{equation*}
ii.

If $f$ has a pole of order $2$ at $z_0\text{,}$ then

\begin{equation*} \mathrm{Res}[f,z_0] =\lim\limits_{z \to z_0}\frac{d}{dz}(z-z_0)^2f(z)\text{.} \end{equation*}
iii.

If $f$ has a pole of order $k$ at $z_0\text{,}$ then

\begin{equation*} \mathrm{Res}[f,z_0] =\frac{1}{(k-1)!}\lim_{z \to z_0}\frac{d^{k-1}}{dz^{k-1}}(z-z_0)^kf(z)\text{.} \end{equation*}

Proof.

If $f$ has a simple pole at $z_0\text{,}$ then the Laurent series is

\begin{equation*} f(z) =\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots\text{.} \end{equation*}

If we multiply both sides of this equation by $(z-z_0)$ and take the limit as $z \to z_0\text{,}$ we obtain

\begin{align*} \lim_{z \to z_0}(z-z_0) f(z) \amp = \lim_{z \to z_0}[a_{-1}+a_0(z-z_0)+a_1(z-z_0)^2+\cdots]\\ \amp =a_{-1} = \mathrm{Res}[f,z_0]\text{.} \end{align*}

which establishes part (i). We proceed to part (iii), as part (ii) is a special case of it. Suppose that $f$ has a pole of order $k$ at $z_0\text{.}$ Then $f$ can be written as

\begin{equation*} f(z) = \frac{a_{-k}}{(z-z_0)^k}+\frac{a_{-k+1}}{(z-z_0)^{k-1}}+\cdots +\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots\text{.} \end{equation*}

Multiplying both sides of this equation by $(z-z_0)^k$ gives

\begin{equation*} (z-z_0)^kf(z) =a_{-k}+\cdots +a_{-1}(z-z_0)^{k-1}+a_0(z-z_0)^k+\cdots\text{.} \end{equation*}

If we differentiate both sides $k-1$ times we get

\begin{align*} \frac{d^{k-1}}{dz^{k-1}}[(z-z_0)^kf(z)] \amp =(k-1) !a_{-1}+k!a_0(z-z_0)\\ \amp + {(k+1)!}\,a_1(z-z_0)^2+\cdots\text{.} \end{align*}

and when we let $z \to z_0$ the result is

\begin{equation*} \lim_{z \to z_0}\frac{d^{k-1}}{dz^{k-1}}[(z-z_0)^kf(z) ] = (k-1)!a_{-1}=(k-1) !\mathrm{Res}[f,z_0]\text{,} \end{equation*}

which establishes part (iii).

Example 8.1.8.

Find the residue of $f(z) =\frac{\pi \cot (\pi z)}{z^2}$ at $z_0=0\text{.}$

Solution.

We write $f(z) =\frac{\pi \cos (\pi z)}{z^2\sin (\pi z)}\text{.}$ Because $z^2\sin \pi z$ has a zero of order $3$ at $z_0=0$ and $\pi \cos (\pi z_0) \ne 0\text{,}$ $f$ has a pole of order $3$ at $z_0\text{.}$ By part (iii) of Theorem 8.2, we have

\begin{align*} \mathrm{Res}[f,0] \amp = \frac{1}{2!}\lim\limits_{z \to 0}\frac{d^2}{dz^2}\pi z\cot (\pi z)\\ \amp = \frac{1}{2}\lim\limits_{z \to 0}\frac{d}{dz}[\pi \cot (\pi z) -\pi^2z\csc^2(\pi z)]\\ \amp = \frac{1}{2}\lim\limits_{z \to 0}[-\pi^2\csc^2(\pi z)-\pi^2\{\csc^2(\pi z) -2\pi z\csc^2(\pi z)\cot(\pi z)\}]\\ \amp = \pi^2\lim\limits_{z \to 0}(\pi z\cot (\pi z)-1) \csc^2(\pi z)\\ \amp = \pi^2\lim\limits_{z \to 0}\frac{\pi z\cos (\pi z) -\sin (\pi z)}{\sin^3(\pi z)}\text{.} \end{align*}

This last limit involves an indeterminate form, which we evaluate by using L’Hôpital’s rule:

\begin{align*} \mathrm{Res}[f,0] \amp = \pi^2\lim\limits_{z \to 0}\frac{-\pi^2z\sin (\pi z)+\pi \cos (\pi z)-\pi \cos(\pi z)}{3\pi \sin^2(\pi z)\cos (\pi z)}\\ \amp = \pi^2\lim\limits_{z \to 0}\frac{-\pi z}{3\sin (\pi z) \cos(\pi z)}\\ \amp = -\frac{\pi^2}{3}\lim\limits_{z \to 0}\frac{\pi z}{\sin(\pi z)}\lim\limits_{z \to 0}\frac{1}{\cos (\pi z)}\\ \amp = -\frac{\pi^2}{3}\text{.} \end{align*}

Example 8.1.9.

Find $\int\limits_{C_3^{+}(0)}\frac{1}{z^4+z^3-2z^2}\,dz\text{.}$

Solution.

We write the integrand as $f(z) =\frac{1}{z^2(z+2)(z-1)}\text{.}$ The singularities of $f$ that lie inside $C_3(0)$ are simple poles at the points $1$ and $-2\text{,}$ and a pole of order $2$ at the origin. We compute the residues as follows:

\begin{align*} \mathrm{Res}[f,0] \amp = \lim_{z \to 0}\frac{d}{dz}[z^2f(z)] = \lim_{z \to 0}\frac{-2z-1}{(z^2+z-2)^2} = -\frac{1}{4}.\\ \mathrm{Res}[f,1] \amp = \lim_{z \to 1}(z-1)f(z) = \lim_{z \to 1}\frac{1}{z^2(z+2)} = \frac{1}{3}, \text{ and }\\ \mathrm{Res}[f,-2] \amp = \lim_{z \to -2}(z+2)f(z) = \lim_{z \to -2}\frac{1}{z^2(z-1)} = -\frac{1}{12}\text{.} \end{align*}

Finally, the residue theorem yields

\begin{equation*} \int\limits_{C_3^{+}(0)}\frac{1}{z^4+z^3-2z^2}\,dz=2\pi i \left[-\frac{1}{4}+\frac{1}{3}-\frac{1}{12}\right] = 0\text{.} \end{equation*}

The answer $\int\limits_{C_3^{+}(0)}\frac{1}{z^4+z^3-2z^2}\,dz=0$ is not at all obvious, and all the preceding calculations are required to get it.

Example 8.1.10.

Find$\int\limits_{C_2^{+}(1)}(z^4+4)^{-1}\,dz\text{.}$

Solution.

The singularities of the integrand $f(z) =\displaystyle\frac{1}{z^4+4}$ that lie inside $C_2(1)$ are simple poles occurring at the points $1\pm i\text{,}$ as the points $-1\pm i$ lie outside $C_2(1)\text{.}$ Factoring the denominator is tedious, so we use a different approach. If $z_0$ is any one of the singularities of $f\text{,}$ then we can use L’Hôpital’s rule to compute $\mathrm{Res}[f,z_0]\text{:}$

\begin{equation*} \mathrm{Res}[f,z_0] = \lim_{z \to z_0}\frac{z-z_0}{z^4+4} = \lim_{z \to z_0}\frac{1}{4z^3} = \frac{1}{4z_0^3} \end{equation*}

. Since $z_0^4=-4\text{,}$ we can simplify this expression further to yield $\mathrm{Res}[f,z_0] = -\frac{1}{16}z_0\text{.}$ Hence $\mathrm{Res}[f,1+i] = \frac{-1-i}{16}\text{,}$ and $\mathrm{Res}[f,1-i] =\frac{-1+i}{16}\text{.}$ We now use the residue theorem to get

\begin{equation*} \int\limits_{C_2^{+}(1)}\frac{1}{z^4+4}\,dz = 2\pi i\left(\frac{-1-i}{16}+\frac{-1+i}{16}\right) = -\frac{\pi i}{4} \end{equation*}

The theory of residues can be used to expand the quotient of two polynomials into its partial fraction representation.

Example 8.1.11.

Let $P(z)$ be a polynomial of degree at most $2\text{.}$ Show that if $a, \, b\text{,}$ and $c$ are distinct complex numbers, then

\begin{align*} f(z) \amp = \frac{P(z)}{(z-a)(z-b)(z-c)}\\ \amp = \frac{A}{z-a}+\frac{B}{z-b}+\frac{C}{z-c}, \text{ where }\\ A \amp = \mathrm{Res}[f,a] =\frac{P(a)}{(a-b)(a-c)}.\\ B \amp = \mathrm{Res}[f,b] =\frac{P(b)}{(b-a)(b-c)} \text{ and }\\ C \amp = \mathrm{Res}[f,c] =\frac{P(c)}{(c-a)(c-b)}\text{.} \end{align*}

Solution.

It will suffice to prove that $A= \frac{B}{z-b} = \frac{-B}{b-a}\frac{1}{1-\frac{z-a}{b-a}} = -\sum\limits_{n=0}^{\infty}\frac{B}{(b-a)^{n+1}}(z-a)^n \frac{C}{z-c}=-\sum\limits_{n=0}^{\infty}\frac{C}{(c-a)^{n+1}}(z-a)^n f(z) = \frac{A}{z-a} - \sum_{n=0}^{\infty}\left[\frac{B}{(b-a)^{n+1}}+\frac{C}{(c-a)^{n+1}}\right](z-a)^n \mathrm{Res}[f,a] =A=\lim\limits_{z \to a}\frac{P(z)}{(z-b)(z-c)} = \frac{P(a)}{(a-b)(a-c)}$$.

Example 8.1.12.

Express $f(z) =\frac{3z+2}{z(z-1)(z-2)}$ in partial fractions.

Solution.

Computing the residues, we obtain

\begin{equation*} \mathrm{Res}[f,0] =1\text{ ,\qquad Res } [f,1] = -5, \text{ and } \mathrm{Res}[f,2]=4\text{.} \end{equation*}

Example 8.1.11 gives us

\begin{equation*} \frac{3z+2}{z(z-1)(z-2)} = \frac{1}{z} - \frac{5}{z-1} + \frac{4}{z-2} \end{equation*}

Remark 8.1.13.

If a repeated root occurs, then the process is similar, and we can easily show that if $P(z)$ has degree of at most $2\text{,}$ then

\begin{equation*} f(z) = \frac{P(z)}{(z-a)^2(z-b)} = \frac{A}{(z-a)^2}+\frac{B}{z-a}+\frac{C}{z-b}\text{,} \end{equation*}

where $A=$ $\mathrm{Res}[(z-a) f(z) ,a]\text{,}$ $B=\mathrm{Res}[f,a]\text{,}$ and $C=\mathrm{Res}[f,b]\text{.}$

Example 8.1.14.

Express $f(z) =\frac{z^2+3z+2}{z^2(z-1)}$ in partial fractions.

Solution.

Using the previous remark, we have

\begin{equation*} f(z) = \frac{A}{(z-a)^2}+\frac{B}{z-a}+\frac{C}{z-b}\text{,} \end{equation*}

where

\begin{align*} A \amp = \mathrm{Res}[zf(z) ,0] = \lim_{z \to 0}\frac{z^2+3z+2}{z-1}=-2.\\ B \amp = \mathrm{Res}[f,0] =\lim_{z \to 0}\frac{d}{dz}\frac{z^2+3z+2}{z-1}\\ \amp = \lim_{z \to 0}\frac{(2z+3) (z-1) -(z^2+3z+2)}{(z-1)^2}=-5, \text{ and }\\ C \amp = \mathrm{Res}[f,1] =\lim\limits_{z \to 1}\frac{z^2+3z+2}{z^2}=6\text{.} \end{align*}

Thus,

\begin{equation*} \frac{z^2+3z+2}{z^2(z-1)} = \frac{-2}{z^2} - \frac{5}{z}+\frac{6}{z-1} \end{equation*}

Exercises Exercises

1.

Find $\mathrm{Res}[f,0]$ for each of the following :

(a)

$f(z) =z^{-1}\exp z\text{.}$

Solution.

$1\text{.}$

(b)

$f(z) =z^{-3}\cosh 4z\text{.}$

(c)

$f(z) =\csc z\text{.}$

Solution.

$1\text{.}$

(d)

$f(z) =\frac{z^2+4z+5}{z^2+z}\text{.}$

(e)

$f(z) =\cot z\text{.}$

Solution.

$1\text{.}$

(f)

$f(z) =z^{-3}\cos z\text{.}$

(g)

$f(z) =z^{-1}\sin z\text{.}$

Solution.

$0\text{.}$

(h)

$f(z) =\frac{z^2+4z+5}{z^3}\text{.}$

(i)

$f(z) =\exp (1+\frac{1}{z})\text{.}$

Solution.

$e\text{.}$

(j)

$f(z) =z^4\sin (\frac{1}{z})\text{.}$

(k)

$f(z) =z^{-1}\csc z\text{.}$

Solution.

$0\text{.}$

(l)

$f(z) =z^{-2}\csc z\text{.}$

(m)

$f(z) =\frac{\exp (4z)-1}{\sin^2z}\text{.}$

Solution.

$4\text{.}$

(n)

$f(z) =z^{-1}\csc^2z\text{.}$

2.

Let $f$ and $g$ have an isolated singularity at $z_0\text{.}$ Show that

\begin{equation} \mathrm{Res}[{f+g},{z_0}] = \mathrm{Res}[f,z_0] + \mathrm{Res}[g,z_0]\text{.}\tag{8.1.3} \end{equation}

3.

Evaluate the following :

(a)

$\int\limits_{C_1^{+}(-1+i)}\frac{1}{z^4+4}\,dz\text{.}$

Solution.

$\frac{\pi +i\pi}{8}\text{.}$

(b)

$\int\limits_{C_2^{+}(i)}\frac{1}{z(z^2-2z+2)}\,dz\text{.}$

(c)

$\int\limits_{C_2^{+}(0)}\frac{\exp z}{z^3+z}\,dz\text{.}$

Solution.

$(1-\cos 1)2\pi i\text{.}$

(d)

$\int\limits_{C_2^{+}(0)}\frac{\sin z}{4z^2-\pi^2}\,dz\text{.}$

(e)

$\int\limits_{C_2^{+}(0)}\frac{\sin z}{z^2+1}\,dz\text{.}$

Solution.

$i2\pi \sinh 1\text{.}$

(f)

$\int\limits_{C_1^{+}(0)}\frac{1}{z^2\sin z}\,dz\text{.}$

(g)

$\int\limits_{C_1^{+}(0)}\frac{1}{z\sin^2z}\,dz\text{.}$

Solution.

$\frac{2\pi i}{3}\text{.}$

4.

Let $f$ and $g$ be analytic at $z_0\text{.}$ If $f(z_0) \ne 0$ and $g$ has a simple zero at $z_0\text{,}$ then show that $\mathrm{Res}\left[\frac{f}{g},z_0\right] = \frac{f(z_0)}{g'(z_0)}\text{.}$

5.

Find $\int\limits_{C}(z-1)^{-2}(z^2+4)^{-1}\,dz$ when

(a)

$C=C_1^{+}(1)\text{.}$

Solution.

$-\frac{4\pi i}{25}\text{.}$

(b)

$C=C_4^{+}(0)\text{.}$

6.

Find $\int\limits_{C}(z^{6}+1)^{-1}dz$ when

(a)

$C=C_{\frac{1}{2}}^{+}(i)\text{.}$

(b)

$C=C_1^{+}(\frac{1+i}{2})\text{.}$

\hint{If $z_0$ is a singularity of $f(z) =\frac{1}{z^{6}+1}\text{,}$ show that $\mathrm{Res}[f,z_0] = -\frac{1}{6}z_0\text{.}$}

7.

Find $\int\limits_{C}(3z^4+10z^2+3)^{-1}\,dz$ when

(a)

$C=C_1^{+}(i\sqrt{3})\text{.}$

Solution.

$-\frac{\pi \sqrt{3}}{24}\text{.}$

(b)

$C=C_1^{+}(\frac{i}{\sqrt{3}})\text{.}$

8.

Find $\int\limits_{C}(z^4-z^3-2z^2)^{-1}dz$ when

(a)

$C=C_{\frac{1}{2}}^{+}(0)\text{.}$

(b)

$C=C_{\frac{3}{2}}^{+}(0)\text{.}$

9.

Use residues to find the partial fraction representations of: \label {8.1.9}

(a)

$\frac{1}{z^2+3z+2}\text{.}$

Solution.

$\frac{1}{z+1}-\frac{1}{z+2}\text{.}$

(b)

$\frac{3z-3}{z^2-z-2}\text{.}$

(c)

$\frac{z^2-7z+4}{z^2(z+4)}\text{.}$

Solution.

$\frac{1}{z^2}-\frac{2}{z}+\frac{3}{z+4}\text{.}$

(d)

$\frac{10z}{(z^2+4)(z^2+9)}\text{.}$

(e)

$\frac{2z^2-3z-1}{(z-1)^3}\text{.}$

Solution.

$\frac{2}{z-1}+\frac{1}{(z-1)^2}-\frac{2}{(z-1)^3}\text{.}$

(f)

$\frac{z^3+3z^2-z+1}{z(z+1)^2(z^2+1)}\text{.}$

10.

Let $f$ be analytic in a simply connected domain $D\text{,}$ and let $C$ be a simple closed positively oriented contour in $D\text{.}$ If $z_0$ is the only zero of $f$ in $D$ and $z_0$ lies interior to $C\text{,}$ then show that $\frac{1}{2\pi i}\int\limits_{C}\frac{f\,'(z)}{f(z)}\,dz = k\text{,}$ where $k$ is the order of the zero at $z_0\text{.}$

11.

Let $f$ be analytic at the points $0,\pm 1,\pm 2,\ldots$ . If $g(z) = \pi f(z) \cot \pi z\text{,}$ then show that $\mathrm{Res}[g,n] = f(n)$ for $n=0, \, \pm 1, \, \pm 2, \, \ldots$ .

Solution.

By Theorem 8.1.7 we have $\mathrm{Res}[g, \, n] = \lim\limits_{z \to n}(z-n)g(z)\text{,}$ where $n$ is any integer. Since $g(z) =\pi f(z) \cot \pi z = \pi f(z)\frac{\cos(\pi z)}{\sin(\pi z)}\text{,}$ and because $f\text{,}$ is analytic at $n\text{,}$ we use L’Hôpital’s rule to get $\lim\limits_{z \to n}\frac{z-n}{\sin (\pi z) }=1\text{.}$ Explain how this gives the result.

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