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Complex Analysis

Russell W. Howell, John H. Mathews

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.
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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*}
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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, \end{equation*}
and \(\mathrm{Res}[f,0] =a_{-1}=2\text{.}\)
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Example 8.1.3.

Evaluate \(\mathrm{Res}[g,0]\) if \(g(z)=\frac{3}{2z+z^2-z^3}\text{.}\)
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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, \(\mathrm{Res}[g,0] =a_{-1}=\frac{3}{2}\text{.}\)
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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}
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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.
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Example 8.1.4.

Evaluate \(\int\limits_{C_1^{+}(0)}\exp(\frac{2}{z})\,dz\text{.}\)
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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*}
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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
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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), 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.
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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.
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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 \(z_0\text{.}\)
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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.
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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 (1). We proceed to part (3), as part (2) 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 (3).
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Example 8.1.8.

Find the residue of \(f(z) =\frac{\pi \cot (\pi z)}{z^2}\) at \(z_0=0\text{.}\)
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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*}
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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*}
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Example 8.1.9.

Evaluate \(\int\limits_{C_3^{+}(0)}\frac{1}{z^4+z^3-2z^2}\,dz\text{.}\)
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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*}
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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*}
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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.
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Example 8.1.10.

Find\(\int\limits_{C_2^{+}(1)}(z^4+4)^{-1}\,dz\text{.}\)
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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 \(\mathrm{Res}[f,z_0]\text{:}\)
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\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*}
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The theory of residues can be used to expand the quotient of two polynomials into its partial fraction representation.
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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}, \quad \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)}, \quad \text{and}\\ C \amp = \mathrm{Res}[f,c] =\frac{P(c)}{(c-a)(c-b)}. \end{align*}
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Solution.
It will suffice to prove that \(A=\mathrm{Res}[f,a]\text{.}\) To do so, expand \(f\) in its Laurent series about the point \(a\) by writing the three terms \(\frac{A}{z-a}, \, \frac{B}{z-b}\) and \(\frac{C}{z-c}\) in their Laurent series about the point \(a\) and adding them.
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The term \(\frac{A}{z-a}\) is itself a one-term Laurent series about the point \(a\text{.}\) The term \(\frac{B}{z-b}\) is analytic at the point \(a\text{,}\) and its Laurent series is actually a Taylor series given by
\begin{equation*} \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, \end{equation*}
which is valid for \(|z-a| \lt |b-a|\text{.}\)
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Likewise, the expansion of the term \(\frac{C}{z-c}\) is
\begin{equation*} \frac{C}{z-c}=-\sum\limits_{n=0}^{\infty}\frac{C}{(c-a)^{n+1}}(z-a)^n, \end{equation*}
which is valid for \(|z-a| \lt |c-a|\text{.}\)
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Thus, the Laurent series of \(f\) about the point \(a\) is
\begin{equation*} 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, \end{equation*}
which is valid for \(|z-a| \lt R\text{,}\) where \(R=\min \{ |b-a| ,\,|c-a| \}\text{.}\)
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Therefore, \(A=\mathrm{Res}[f,a]\text{,}\) and calculation reveals that
\begin{equation*} \mathrm{Res}[f,a] =A=\lim\limits_{z \to a}\frac{P(z)}{(z-b)(z-c)} = \frac{P(a)}{(a-b)(a-c)}. \end{equation*}
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Example 8.1.12.

Express \(f(z) =\frac{3z+2}{z(z-1)(z-2)}\) in partial fractions.
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Solution.
Computing the residues, we obtain
\begin{equation*} \mathrm{Res}[f,0] =1, \quad \mathrm{Res}[f,1] = -5, \quad \text{and} \quad \mathrm{Res}[f,2]=4. \end{equation*}
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Example 8.1.11 then 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*}
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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}, \end{equation*}
where \(A = \mathrm{Res}[(z-a) f(z) ,a], \quad B=\mathrm{Res}[f,a], \quad \text{and} \quad C=\mathrm{Res}[f,b]\text{.}\)
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Example 8.1.14.

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

1.

Find \(\mathrm{Res}[f,0]\) for each of the following:
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(b)
\(f(z) =z^{-3}\cosh 4z\text{;}\)
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(d)
\(f(z) =\frac{z^2+4z+5}{z^2+z}\text{;}\)
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(f)
\(f(z) =z^{-3}\cos z\text{;}\)
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(h)
\(f(z) =\frac{z^2+4z+5}{z^3}\text{;}\)
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(i)
\(f(z) =\exp (1+\frac{1}{z})\text{;}\)
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Solution.
\(e\text{.}\)
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(j)
\(f(z) =z^4\sin (\frac{1}{z})\text{;}\)
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(l)
\(f(z) =z^{-2}\csc z\text{;}\)
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(m)
\(f(z) =\frac{\exp (4z)-1}{\sin^2z}\text{;}\)
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Solution.
\(4\text{.}\)
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(n)
\(f(z) =z^{-1}\csc^2z\text{.}\)
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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}
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3.

Evaluate the following:
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(a)
\(\int\limits_{C_1^{+}(-1+i)}\frac{1}{z^4+4}\,dz\text{;}\)
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Solution.
\(\frac{\pi +i\pi}{8}\text{.}\)
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(b)
\(\int\limits_{C_2^{+}(i)}\frac{1}{z(z^2-2z+2)}\,dz\text{;}\)
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(c)
\(\int\limits_{C_2^{+}(0)}\frac{\exp z}{z^3+z}\,dz\text{;}\)
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Solution.
\((1-\cos 1)2\pi i\text{.}\)
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(d)
\(\int\limits_{C_2^{+}(0)}\frac{\sin z}{4z^2-\pi^2}\,dz\text{;}\)
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(e)
\(\int\limits_{C_2^{+}(0)}\frac{\sin z}{z^2+1}\,dz\text{;}\)
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Solution.
\(i2\pi \sinh 1\text{.}\)
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(f)
\(\int\limits_{C_1^{+}(0)}\frac{1}{z^2\sin z}\,dz\text{;}\)
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(g)
\(\int\limits_{C_1^{+}(0)}\frac{1}{z\sin^2z}\,dz\text{.}\)
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Solution.
\(\frac{2\pi i}{3}\text{.}\)
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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 \(\mathrm{Res}\left[\frac{f}{g},z_0\right] = \frac{f(z_0)}{g'(z_0)}\text{.}\)
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5.

Find \(\int\limits_{C}(z-1)^{-2}(z^2+4)^{-1}\,dz\) when:
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(a)
\(C=C_1^{+}(1)\text{;}\)
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Solution.
\(-\frac{4\pi i}{25}\text{.}\)
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6.

Find \(\int\limits_{C}(z^{6}+1)^{-1}dz\) when:
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(a)
\(C=C_{\frac{1}{2}}^{+}(i)\text{;}\)
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(b)
\(C=C_1^{+}(\frac{1+i}{2})\text{.}\)
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Hint.
If \(\, z_0 \,\) is a singularity of \(\,f(z) =\frac{1}{z^{6}+1}, \,\) show that \(\, \mathrm{Res}[f,z_0] = -\frac{1}{6}z_0.\)
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7.

Find \(\int\limits_{C}(3z^4+10z^2+3)^{-1}\,dz\) when:
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(a)
\(C=C_1^{+}(i\sqrt{3})\text{;}\)
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Solution.
\(-\frac{\pi \sqrt{3}}{24}\text{.}\)
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(b)
\(C=C_1^{+}(\frac{i}{\sqrt{3}})\text{.}\)
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8.

Find \(\int\limits_{C}(z^4-z^3-2z^2)^{-1}dz\) when:
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(a)
\(C=C_{\frac{1}{2}}^{+}(0)\text{;}\)
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(b)
\(C=C_{\frac{3}{2}}^{+}(0)\text{.}\)
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9.

Use residues to find the partial fraction representations of:
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(a)
\(\frac{1}{z^2+3z+2}\text{;}\)
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Solution.
\(\frac{1}{z+1}-\frac{1}{z+2}\text{.}\)
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(b)
\(\frac{3z-3}{z^2-z-2}\text{;}\)
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(c)
\(\frac{z^2-7z+4}{z^2(z+4)}\text{;}\)
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Solution.
\(\frac{1}{z^2}-\frac{2}{z}+\frac{3}{z+4}\text{.}\)
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(d)
\(\frac{10z}{(z^2+4)(z^2+9)}\text{;}\)
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(e)
\(\frac{2z^2-3z-1}{(z-1)^3}\text{;}\)
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Solution.
\(\frac{2}{z-1}+\frac{1}{(z-1)^2}-\frac{2}{(z-1)^3}\text{.}\)
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(f)
\(\frac{z^3+3z^2-z+1}{z(z+1)^2(z^2+1)}\text{.}\)
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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 \(C\text{,}\) show that
\begin{equation*} \frac{1}{2\pi i}\int\limits_{C}\frac{f\,'(z)}{f(z)}\,dz = k, \end{equation*}
where \(k\) is the order of the zero at \(z_0\text{.}\)
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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 \(n=0, \, \pm 1, \, \pm 2, \, \ldots\) .
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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 calculation gives the result.
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