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\[b_{1}=\operatorname{Res}(f(z),a)=\frac{1}{(k-1)!}\lim_{z\to a}\left[\frac{d^{ k-1}}{dz^{k-1}}[(z-a)^{k}f(z)]\right].\qed\]
**Remark 79**.: Let \(f(z)=\frac{g(z)}{h(z)}\) where \(g\) and \(h\) are holomorphic at \(z=a\). Furthermore, suppose \(g(a)\neq 0\) and \(h(a)=0\), but \(h^{\prime}(a)\neq 0\) ( in such case we say \(h\) has a zero of order \(1\) at \(a\)). Then \(f\) has a simple pole at \(z=a\), and
\[\operatorname{Res}(f(z),a)=\frac{g(a)}{h^{\prime}(a)}.\]
This easy way of finding residues for such functions \(f(z)=\frac{g(z)}{h(z)}\) follows from the fact that
\[\operatorname{Res}(f(z),a)=\lim_{z\to a}(z-a)\frac{g(z)}{h(z)}=\lim_{z\to a} \frac{g(z)}{\frac{h(z)}{z-a}}=\frac{g(a)}{h^{\prime}(a)}.\]
**Example 116**.:
1. \(f(z)=\frac{1}{1+z^{2}}=\frac{1}{(z+i)(z-i)}\) has simple poles at \(z=i\) and \(z=-i\). Find the residues there. _Solution._ To find the residues, we calculate \[\operatorname{Res}(f(z),i)=\lim_{z\to i}(z-i)f(z)=\lim_{z\to i}(z-i)\frac{1}{( z+i)(z-i)}=\frac{1}{2i}.\] and \[\operatorname{Res}(f(z),-i)=\lim_{z\to-i}(z+i)f(z)=\lim_{z\to-i}(z+i)\frac{1}{( z+i)(z-i)}=-\frac{1}{2i}.\] The residues at \(z=i\) and \(z=-i\) are \(\frac{1}{2i}\) and \(-\frac{1}{2i}\), respectively.
* \(f(z)=\frac{e^{z}}{z^{3}}\) has a pole of order \(3\) at \(z=0\) since \(\lim_{z\to 0}z^{3}\cdot\frac{e^{z}}{z^{3}}=1\neq 0\). Find the residues at the pole. _Solution._ To find the residue, we calculate \[\text{Res}(f(z),0)=\frac{1}{2!}\lim_{z\to 0}\left[\frac{d^{2}}{dz^{2}}(z-0)^{3}f(z) \right]=\frac{1}{2!}\lim_{z\to 0}\left[\frac{d^{2}}{dz^{2}}z^{3}\cdot\frac{e^{z}}{z^{3}} \right]=\frac{1}{2}.\] Thus the residue of the pole at \(z=0\) is \(\frac{1}{2}\).
* Calculate the residues of the function \(f(z)=\frac{\sin z}{(z-i)(z+2)^{2}}\) at its singularities. _Solution._ Observe that \(z=i\) and \(z=-2\) are singularities for \(f(z)\), and that \(z=i\) is a simple pole while \(z=-2\) is a pole of order \(2\). Therefore, \[\text{Res}(f(z),i)=\lim_{z\to i}(z-i)\frac{\sin z}{(z-i)(z+2)^{2}}=\frac{\sin i }{(i+2)^{2}}\] and \[\text{Res}(f(z),-2)=\lim_{z\to-2}\frac{d}{dz}\left[(z+2)^{2} \frac{\sin z}{(z-i)(z+2)^{2}}\right]=\lim_{z\to-2}\frac{d}{dz}\left[\frac{\sin z }{(z-i)}\right]\] \[=\lim_{z\to-2}\frac{(z-i)\cos z-\sin z}{(z-i)^{2}}=\frac{(-2-i) \cos(2)+\sin(2)}{3+4i}.\]
* The function \(f(z)=\frac{\sin z-z}{z^{4}}\) has a pole at \(z=0\), but the order of this pole is not immediately evident. However, if we write \(\sin z-z=-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\cdots\) then \[f(z)=\frac{-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\cdots}{z^{4}}=-\frac{1}{3!z}+ \frac{z}{5!}-\cdots.\] Thus we have that \(\text{Res}(f(z),0)=-\frac{1}{3!}=-\frac{1}{6}\).
### Cauchy’s Residue Theorem
**Theorem 148**.: _Let \(f\) be holomorphic inside and on a positively oriented contour except for a finite number of poles \(a_{1},a_{2},\ldots,a_{n}\) inside \(\gamma\) (Figure 3.42). Then_
Proof
: Consider the Laurent expansion at \(a_{k}\):
\[f(z)=\underbrace{\frac{b_{m}}{(z-a_{k})^{m}}+\cdots+\frac{b_{1}}{(z-a_{k})^{m}}}_{ \text{Call this part $f_{k}(z)$}}+\sum_{n=0}^{\infty}a_{n}(z-a_{k})^{n}.\]
Define a function \(g:=f(z)-\sum_{k=1}^{N}f_{k}(z)\). Then \(g(z)\) has only removable singularities at \(a_{1},a_{2},\ldots,a_{N}.\) Remove them and use Cauchy’s theorem to deduce that \(\int_{\gamma}g(z)dz=0\). But
\[\int_{\gamma}g(z)dz=\int_{\gamma}f(z)dz-\sum_{k=1}^{N}\int_{\gamma}f_{k}(z)dz=0.\]
Hence,
\[\int_{\gamma}f(z)dz=\sum_{k=1}^{N}\int_{\gamma}f_{k}(z)\,dz.\]
Note that the integral around a pole can be found easily using the fundamental integral and we have
\[\int_{\gamma}f_{k}(z)dz=2\pi i\text{Res}(f(z),a_{k})\]
(see Exercise 17 below). Therefore,
\[\int_{\gamma}f(z)dz=2\pi i\sum_{k=1}^{N}\text{Res}(f(z),a_{k}).\qed\]
Figure 3.42: \(f\) is not holomorphic for finite number of poles \(a_{1},a_{2},a_{3},a_{4}\) in \(\gamma\)
[MISSING_PAGE_EMPTY:703]
Only \(z_{1}\) and \(z_{2}\) are inside the given contour. Therefore,
\[\int_{\gamma}\frac{dz}{1+z^{4}}=2\pi i\left[\mathrm{Res}(f(z),z_{1})+\mathrm{Res} (f(z),z_{2})\right].\]
Now
\[\mathrm{Res}(f(z),z_{1})=\lim_{z\to e^{i\pi/4}}\frac{z-e^{\pi/4}}{1+z^{4}}= \lim_{z\to e^{i\pi/4}}\frac{1}{4z^{3}}=\frac{1}{4}e^{-3\pi i/4}\quad\text{and}\]
\[\mathrm{Res}(f(z),z_{2})=\lim_{z\to e^{3\pi i/4}}\frac{z-e^{3\pi i/4}}{1+z^{4} }=\lim_{z\to e^{3\pi i/4}}\frac{1}{4z^{3}}=\frac{1}{4}e^{-\pi i/4}.\]
Therefore
\[\int_{\gamma}\frac{dz}{1+z^{4}}=2\pi i\left[\frac{1}{4}e^{-3\pi i/4}+\frac{1}{ 4}e^{-\pi i/4}\right]=\frac{\pi}{\sqrt{2}}.\]
**Example 3.** Use Cauchy’s Residue Theorem to evaluate the following real trigonometric integral
\[\int_{0}^{2\pi}\frac{\sin^{2}(\theta)}{5+4\cos(\theta)}d\theta.\]
_Solution:_ First we establish the identification
\[z=e^{i\theta}\quad 0\leq\theta\leq 2\pi.\]
With this choice of identification, we have that \(\frac{1}{z}=e^{-i\theta}\). Since
\[\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+\frac{1}{z}}{2}\quad \text{and}\quad\sin\theta=\frac{e^{i\theta-e^{-i\theta}}}{2i}=\frac{z-\frac{1} {z}}{2i},\]
we have that
\[\cos\theta=\frac{1}{2}\left(z+\frac{1}{z}\right)\quad\text{and}\qquad\sin \theta=\frac{1}{2i}\left(z-\frac{1}{z}\right).\]
Furthermore, when integrating along \(\gamma\),
\[z=e^{i\theta}\implies dz=izd\theta\implies\frac{dz}{iz}=d\theta.\]
Secondly, observe that \(5+4\cos\theta\) is never zero. Thus the integrand is finite over \([0,2\pi]\). Performing the above substitutions leads to
\[\int_{0}^{2\pi}\frac{\sin^{2}\theta}{5+4\cos\theta}d\theta=\int_{\gamma}\frac {\left[\frac{1}{2i}\left(z-\frac{1}{z}\right)\right]^{2}}{5+4\cdot\frac{1}{2} \left(z+\frac{1}{z}\right)}\frac{dz}{iz},\]
which can be reduced to \[-\frac{1}{4i}\int_{\gamma}\frac{(z^{2}-1)^{2}}{z^{2}(2z^{2}+5z+2)}dz=-\frac{1}{4i} \int_{\gamma}\frac{(z^{2}-1)^{2}}{2z^{2}\left(z+\frac{1}{2}\right)(z+2)}dz.\]
Let \(g(z)=\frac{(z^{2}-1)^{2}}{2z^{2}\left(z+\frac{1}{2}\right)(z+2)}\). Observe that \(g(z)\) has simple poles at \(z=-\frac{1}{2}\), \(z=-2\) and a pole of order \(2\) at \(z=0\). However, only \(-\frac{1}{2}\) and \(0\) lie inside the unit circle \(\gamma\), so that
\[\int_{0}^{2\pi}\frac{\sin^{2}\theta}{5+4\cos\theta}d\theta=-\frac{1}{4i}\cdot 2 \pi i\left[\operatorname{Res}(g(z),-\frac{1}{2})+\operatorname{Res}(g(z),0) \right].\]
Now we calculate the residues as follows:
\[\operatorname{Res}\left(g(z),-\frac{1}{2}\right)=\lim_{z\to\frac{1}{2}}[(z+ \frac{1}{2})\cdot g(z)]=\lim_{z\to-\frac{1}{2}}\frac{(z^{2}-1)^{2}}{2z^{2}(z+ 2)}=\frac{3}{4}\]
and
\[\operatorname{Res}\left(g(z),0\right)=\lim_{z\to 0}\frac{1}{1!}\frac{d}{dz}[z^{2}g(z)]= \lim_{z\to 0}\frac{d}{dz}\left[\frac{(z-1)^{2}}{2z^{2}+5z+2}\right]\] \[=\lim_{z\to 0}\frac{(2z^{2}+5z+2)\cdot 2(z^{2}-1)2z-(z-1)^{2}(4z+5)} {(2z^{2}+5z+2)^{2}}=-\frac{5}{4}.\]
Hence,
\[\int_{0}^{2\pi}\frac{\sin^{2}\theta}{5+4\cos\theta}d\theta=-\frac{1}{4i}\cdot 2 \pi i\left[\frac{3}{4}-\frac{5}{4}\right]=\frac{\pi}{4}.\]
### Exercises
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