Let C denote the square contour with corners at \pm 2,\pm 2i and which is taken in counter clockwise direction. Use the residue theorem to evaluate the integral \int _{C}f\left ( x\right ) dz for the following functions
(a) \frac{e^{-z}}{z-i\frac{\pi }{2}}, (b) \frac{\cos z}{z\left ( z^{2}+8\right ) } (c) \frac{z}{z+1}
Solution
The function f\left ( z\right ) =\frac{e^{-z}}{z-i\frac{\pi }{2}} has a simple pole at z=i\frac{\pi }{2}\approx 1.57i, hence it is inside the contour.
Hence by residue theorem{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\ \operatorname{Residue}\left ( f\left ( z\right ) \right ) _{z=z_{0}} So we just need to find the residue of f\left ( z\right ) at z=z_{0}=i\frac{\pi }{2}. Since this is a simple plot, then the residue is given by\begin{align*} \operatorname{Residue}\left ( f\left ( z\right ) \right ) _{z=z_{0}} & =\lim _{z\rightarrow z_{0}}\left ( z-z_{0}\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow i\frac{\pi }{2}}\left ( z-i\frac{\pi }{2}\right ) \frac{e^{-z}}{z-i\frac{\pi }{2}}\\ & =\lim _{z\rightarrow i\frac{\pi }{2}}e^{-z}\\ & =e^{-i\frac{\pi }{2}}\\ & =\cos \left ( \frac{\pi }{2}\right ) -i\sin \left ( \frac{\pi }{2}\right ) \\ & =-i \end{align*}
Therefore \begin{align*}{\displaystyle \oint \limits _{C}} \frac{e^{-z}}{z-i\frac{\pi }{2}}dz & =2\pi i\left ( -i\right ) \\ & =2\pi \end{align*}
The function f\left ( z\right ) =\frac{\cos z}{z\left ( z^{2}+8\right ) } has one simple pole at z=0 which is inside the contour, and a poles at z=\pm i\sqrt{8}=\pm 2i\sqrt{2}\approx \pm 2.83i but these are outside the contour.
Therefore by residue theorem, only the pole inside the contour which is at z=0 will contribute to the integral. So we just need to find residue at z=0\begin{align*} \operatorname{Residue}\left ( f\left ( z\right ) \right ) _{z=z_{0}} & =\lim _{z\rightarrow z_{0}}\left ( z-z_{0}\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow 0}\left ( z\right ) \frac{\cos z}{z\left ( z^{2}+8\right ) }\\ & =\lim _{z\rightarrow 0}\frac{\cos z}{\left ( z^{2}+8\right ) }\\ & =\frac{1}{8} \end{align*}
Therefore\begin{align*}{\displaystyle \oint \limits _{C}} \frac{\cos z}{z\left ( z^{2}+8\right ) }dz & =2\pi i\left ( \frac{1}{8}\right ) \\ & =i\frac{\pi }{4} \end{align*}
The function f\left ( z\right ) =\frac{z}{2\left ( z+1\right ) } has one simple pole at z=-1 which is inside the contour.
So we just need to find residue at z=-1\begin{align*} \operatorname{Residue}\left ( f\left ( z\right ) \right ) _{z=z_{0}} & =\lim _{z\rightarrow z_{0}}\left ( z-z_{0}\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow -1}\left ( z+1\right ) \frac{z}{2\left ( z+1\right ) }\\ & =\lim _{z\rightarrow -1}\frac{z}{2}\\ & =\frac{-1}{2} \end{align*}
Therefore\begin{align*}{\displaystyle \oint \limits _{C}} \frac{z}{2\left ( z+1\right ) }dz & =2\pi i\left ( \frac{-1}{2}\right ) \\ & =-i\pi \end{align*}
Assume that f\left ( z\right ) is analytic on and interior to a closed contour C and that the point z_{0} lies inside C. Show that{\displaystyle \oint \limits _{C}} \frac{f^{\prime }\left ( z\right ) }{z-z_{0}}dz={\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{2}}dz Solution
We see that g\left ( z\right ) =\frac{f^{\prime }\left ( z\right ) }{z-z_{0}} has a simple pole at z=z_{0}. Therefore\begin{equation}{\displaystyle \oint \limits _{C}} g\left ( z\right ) dz=2\pi i\left ( b_{1}\right ) \tag{1} \end{equation} Where b_{1} is the \operatorname{Residue} of g\left ( z\right ) at z_{0}. By definition the residue of a simple pole is found as follows\begin{align*} b_{1} & =\lim _{z\rightarrow z_{0}}\left ( z-z_{0}\right ) g\left ( z\right ) \\ & =\lim _{z\rightarrow z_{0}}\left ( z-z_{0}\right ) \frac{f^{\prime }\left ( z\right ) }{z-z_{0}}\\ & =\lim _{z\rightarrow z_{0}}f^{\prime }\left ( z\right ) \\ & =f^{\prime }\left ( z_{0}\right ) \end{align*}
Hence (1) becomes\begin{align}{\displaystyle \oint \limits _{C}} g\left ( z\right ) dz & =\left ( 2\pi i\right ) f^{\prime }\left ( z_{0}\right ) \nonumber \\{\displaystyle \oint \limits _{C}} \frac{f^{\prime }\left ( z\right ) }{z-z_{0}}dz & =\left ( 2\pi i\right ) f^{\prime }\left ( z_{0}\right ) \tag{2} \end{align}
But per lecture notes, page 46 on complex analysis, it shows that f^{\prime }\left ( z_{0}\right ) =\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{2}}dz Substituting the above back into RHS of (2) results in{\displaystyle \oint \limits _{C}} \frac{f^{\prime }\left ( z\right ) }{z-z_{0}}dz=2\pi i\left ( \frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{2}}dz\right ) Therefore{\displaystyle \oint \limits _{C}} \frac{f^{\prime }\left ( z\right ) }{z-z_{0}}dz={\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{2}}dz QED.
Give the Laurent series expansion both in powers of z and in powers of \left ( z-1\right ) for the function \frac{1}{z^{2}\left ( 1-z\right ) }
Solution
There is a pole of order 2 at z=0 and a pole of order one at z=1. Therefore, there is a Laurent series expansion about z=0 which is valid in inside a disk or radius 1 centered at z=0. Around z=1 there is another Laurent series expansion of the function, which is valid inside a disk centered at z=0 of radius 1.
Laurent series expansion around z=0\begin{align*} \frac{1}{z^{2}\left ( 1-z\right ) } & =\frac{1}{z^{2}}\frac{1}{\left ( 1-z\right ) }\\ & =\frac{1}{z^{2}}\left ( 1+z+z^{2}+z^{3}+\cdots \right ) \qquad \left \vert z\right \vert <1\\ & =\frac{1}{z^{2}}+\frac{1}{z}+1+z+z^{2}+z^{3}+\cdots \end{align*}
We see from the above that the residue at z=0 is 1 which is the coefficient of \frac{1}{z} term.
Laurent series expansion around z=1
Let u=z-1, hence z=u+1 and the function \frac{1}{z^{2}\left ( 1-z\right ) } in terms of u becomes\begin{equation} \frac{1}{\left ( 1+u\right ) ^{2}\left ( -u\right ) }=\frac{-1}{u}\frac{1}{\left ( 1+u\right ) ^{2}} \tag{1} \end{equation} But \frac{1}{\left ( 1+u\right ) ^{2}}=\left ( 1+u\right ) ^{-2}. Applying Binomial expansion \left ( 1+x\right ) ^{n}=1+nx+\frac{n\left ( n-1\right ) }{2!}x^{2}+\frac{n\left ( n-1\right ) \left ( n-2\right ) }{3!}x^{3}+\cdots which is valid for \left \vert x\right \vert <1 then we see that for n=-2 we obtain \left ( 1+u\right ) ^{-2}=1+\left ( -2\right ) u+\frac{\left ( -2\right ) \left ( -2-1\right ) }{2!}u^{2}+\frac{\left ( -2\right ) \left ( -2-1\right ) \left ( -2-2\right ) }{3!}u^{3}+\cdots The above is valid for \left \vert u\right \vert <1 or \left \vert z-1\right \vert <1 or 0<z<2. Simplifying the above gives \frac{1}{\left ( 1-u\right ) ^{2}}=1-2u+3u^{2}-4u^{3}+\cdots Substituting the above back into (1) gives\begin{align*} \frac{-1}{u}\frac{1}{\left ( 1+u\right ) ^{2}} & =\frac{-1}{u}\left ( 1-2u+3u^{2}-4u^{3}+\cdots \right ) \\ & =\frac{-1}{u}+2-3u+4u^{2}-\cdots \end{align*}
But since u=z-1 then the above becomes \frac{1}{z^{2}\left ( 1-z\right ) }=\frac{-1}{z-1}+2-3\left ( z-1\right ) +4\left ( z-1\right ) ^{2}-5\left ( z-1\right ) ^{3}+\cdots We see from the above that the residue of f\left ( z\right ) is -1 at z=1.
In summary
Note that there is another Laurent series expansions that can be found, which is for the region \,1<\left \vert z\right \vert <\infty , which is outside a disk of radius 1 centered at z=0. But the problem is asking for the above two expansions only.
Evaluate the integral \int _{0}^{\infty }\frac{dx}{1+x^{4}}dx
solution
Since the integrand is even, then I=\frac{1}{2}\int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx Now we consider the following contour
Therefore{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=\left ( \lim _{R\rightarrow \infty }\int _{-R}^{0}f\left ( x\right ) dx+\lim _{\tilde{R}\rightarrow \infty }\int _{0}^{\tilde{R}}f\left ( x\right ) dx\right ) +\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) dz Using Cauchy principal value the integral above can be written as\begin{align*}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz & =\lim _{R\rightarrow \infty }\int _{-R}^{R}f\left ( x\right ) dx+\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) d\\ & =2\pi i\sum \operatorname{Residue} \end{align*}
Where \sum \operatorname{Residue} is sum of residues of \frac{1}{z^{4}+1} for poles that are inside the contour C. Therefore the above becomes\begin{align} \lim _{R\rightarrow \infty }\int _{-R}^{R}f\left ( x\right ) dx & =2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) dz\nonumber \\ \int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx & =2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C_{R}}\frac{1}{z^{4}+1}dz \tag{1} \end{align}
Now we will show that \lim _{R\rightarrow \infty }\int _{C_{R}}\frac{1}{z^{4}+1}dz=0. Since \begin{align} \left \vert \int _{C_{R}}\frac{1}{z^{4}+1}dz\right \vert & \leq ML\nonumber \\ & =\left \vert f\left ( z\right ) \right \vert _{\max }\left ( \pi R\right ) \tag{2} \end{align}
But f\left ( z\right ) =\frac{1}{\left ( z^{2}-i\right ) \left ( z^{2}+i\right ) } Hence, and since z=R\ e^{i\theta } then \left \vert f\left ( z\right ) \right \vert _{\max }\leq \frac{1}{\left \vert z^{2}-i\right \vert _{\min }\left \vert z^{2}+i\right \vert _{\min }} Using the inverse triangle inequality then \left \vert z^{2}-i\right \vert \geq \left \vert z\right \vert ^{2}+1 and \left \vert z^{2}+i\right \vert \geq \left \vert z\right \vert ^{2}-1, and because \left \vert z\right \vert =R then the above becomes\begin{align*} \left \vert f\left ( z\right ) \right \vert _{\max } & \leq \frac{1}{\left ( R^{2}+1\right ) \left ( R^{2}-1\right ) }\\ & =\frac{1}{R^{4}-1} \end{align*}
Therefore (2) becomes \left \vert \int _{C_{R}}\frac{1}{z^{4}+1}dz\right \vert \leq \frac{\pi R}{R^{4}-1} Then it is clear that as R\rightarrow \infty the above goes to zero since \lim _{R\rightarrow \infty }\frac{\pi R}{R^{4}-1}=\lim _{R\rightarrow \infty }\frac{\frac{\pi }{R^{3}}}{1-\frac{1}{R^{4}}}=\frac{0}{1}=0. Equation (1) now simplifies to\begin{equation} \int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx=2\pi i\sum \operatorname{Residue} \tag{2A} \end{equation} We just now need to find the residues of \frac{1}{z^{4}+1} located in upper half plane. The zeros of the denominator z^{4}+1=0 are at z=-1^{\frac{1}{4}}=\left ( e^{i\pi }\right ) ^{\frac{1}{4}}, then the first zero is at e^{i\frac{\pi }{4}}, and the second zero at e^{i\left ( \frac{\pi }{4}+\frac{\pi }{2}\right ) }=e^{i\left ( \frac{3}{4}\pi \right ) } and the third zero at e^{i\left ( \frac{3}{4}\pi +\frac{\pi }{2}\right ) }=e^{i\left ( \frac{5}{4}\pi \right ) } and the fourth zero at e^{i\left ( \frac{5}{4}\pi +\frac{\pi }{2}\right ) }=e^{i\frac{7}{4}\pi }. Hence poles are at\begin{align*} z_{1} & =e^{i\frac{\pi }{4}}\\ z_{2} & =e^{i\frac{3}{4}\pi }\\ z_{3} & =e^{i\frac{5}{4}\pi }\\ z_{4} & =e^{i\frac{7}{4}\pi } \end{align*}
Out of these only the first two are in upper half plane. Hence since these are simple poles, we can use the following to find the residues\begin{align} \operatorname{Residue}\left ( z_{1}\right ) & =\lim _{z\rightarrow z_{1}}\left ( z-z_{1}\right ) f\left ( z\right ) \nonumber \\ & =\lim _{z\rightarrow z_{1}}\left ( z-z_{1}\right ) \frac{1}{z^{4}-1}\nonumber \end{align}
Applying L’Hopitals rule, the above becomes\begin{align*} \operatorname{Residue}\left ( z_{1}\right ) & =\lim _{z\rightarrow z_{1}}\frac{\frac{d}{dz}\left ( z-z_{1}\right ) }{\frac{d}{dz}\left ( z^{4}-1\right ) }\\ & =\lim _{z\rightarrow e^{i\frac{\pi }{4}}}\frac{1}{4z^{3}}\\ & =\frac{1}{4\left ( e^{i\frac{\pi }{4}}\right ) ^{3}}\\ & =\frac{1}{4e^{i\frac{3\pi }{4}}} \end{align*}
Similarly for the other residue\begin{align} \operatorname{Residue}\left ( z_{2}\right ) & =\lim _{z\rightarrow z_{2}}\left ( z-z_{2}\right ) f\left ( z\right ) \nonumber \\ & =\lim _{z\rightarrow z_{2}}\left ( z-z_{2}\right ) \frac{1}{z^{4}-1}\nonumber \end{align}
Applying L’Hopitals\begin{align*} \operatorname{Residue}\left ( z_{2}\right ) & =\lim _{z\rightarrow e^{i\frac{3}{4}\pi }}\frac{1}{4z^{3}}\\ & =\frac{1}{4\left ( e^{i\frac{3}{4}\pi }\right ) ^{3}}\\ & =\frac{1}{4e^{i\frac{9\pi }{4}}}\\ & =\frac{1}{4e^{i\frac{\pi }{4}}} \end{align*}
Now that we found all the residues, then (2A) becomes \begin{align*} \int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx & =2\pi i\left ( \frac{1}{4e^{i\frac{3\pi }{4}}}+\frac{1}{4e^{i\frac{\pi }{4}}}\right ) \\ & =2\pi i\left ( \frac{\sqrt{2}}{4i}\right ) \\ & =\frac{1}{2}\sqrt{2}\pi \end{align*}
But \int _{0}^{\infty }\frac{1}{x^{4}+1}dx=\frac{1}{2}\int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx, therefore \begin{align*} \int _{0}^{\infty }\frac{1}{x^{4}+1}dx & =\frac{1}{2}\left ( \frac{1}{2}\sqrt{2}\pi \right ) \\ & =\frac{1}{4}\sqrt{2}\pi \\ & =\frac{2}{4\sqrt{2}}\pi \\ & =\frac{1}{2\sqrt{2}}\pi \end{align*}
Evaluate the following integral \int _{0}^{2\pi }\frac{\sin ^{2}\theta }{a+b\cos \theta }d\theta when a>\left \vert b\right \vert
solution
This is converted to complex integration by using z=re^{i\theta }=e^{i\theta } since r=1. Therefore dz=ie^{i\theta }d\theta or dz=izd\theta In addition,\begin{align*} \cos \theta & =\frac{e^{i\theta }+e^{-i\theta }}{2}\\ & =\frac{z+z^{-1}}{2} \end{align*}
And\begin{align*} \sin ^{2}\theta & =\frac{1}{2}-\frac{1}{2}\cos 2\theta \\ & =\frac{1}{2}-\frac{1}{2}\left ( \frac{e^{i2\theta }+e^{-i2\theta }}{2}\right ) \\ & =\frac{1}{2}-\frac{1}{2}\left ( \frac{z^{2}+z^{-2}}{2}\right ) \\ & =\frac{1}{2}-\frac{1}{4}\left ( z^{2}+z^{-2}\right ) \end{align*}
Using all of the above back in the original integral gives\begin{align*} I & =\int _{0}^{2\pi }\frac{\sin ^{2}\theta }{a+b\cos \theta }d\theta \\ & ={\displaystyle \oint \limits _{C}} \frac{\frac{1}{2}-\frac{1}{4}\left ( z^{2}+z^{-2}\right ) }{a+b\left ( \frac{z+z^{-1}}{2}\right ) }\frac{dz}{iz} \end{align*}
Where the contour C is around the unit circle in counter clockwise direction. Therefore\begin{align*} I & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{\frac{1}{2}-\frac{1}{4}\left ( z^{2}+\frac{1}{z^{2}}\right ) }{a+\frac{b}{2}\left ( z+\frac{1}{z}\right ) }\frac{dz}{z}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{\frac{1}{2}-\frac{1}{4}\left ( \frac{z^{4}+1}{z^{2}}\right ) }{a+\frac{b}{2}\left ( \frac{z^{2}+1}{z}\right ) }\frac{dz}{z}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{\frac{\frac{z^{2}}{2}-\frac{1}{4}\left ( z^{4}+1\right ) }{z^{2}}}{\frac{az+\frac{b}{2}\left ( z^{2}+1\right ) }{z}}\frac{dz}{z}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{\frac{z^{2}}{2}-\frac{1}{4}\left ( z^{4}+1\right ) }{az+\frac{b}{2}\left ( z^{2}+1\right ) }\frac{dz}{z^{2}}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{\frac{2z^{2}}{4}-\frac{1}{4}\left ( z^{4}+1\right ) }{\frac{2az}{2}+\frac{b}{2}\left ( z^{2}+1\right ) }\frac{dz}{z^{2}}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{2z^{2}-z^{4}-1}{4az+2bz^{2}+2b}\frac{dz}{z^{2}}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{2bz^{2}+4az+2b}dz\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{1}{z^{2}}\frac{\frac{1}{b}z^{2}-\frac{1}{2b}z^{4}-\frac{1}{2b}}{z^{2}+\frac{2a}{b}z+1}dz\\ & =\frac{1}{2bi}{\displaystyle \oint \limits _{C}} \frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z^{2}+\frac{2a}{b}z+1\right ) }dz \end{align*}
Now we can use the residue theorem. There is a pole at z=0 of order 2 and two poles which are the roots of z^{2}+\frac{2a}{b}z+1=0. Hence I=2\pi i\sum \operatorname{Residue} First we find the roots of z^{2}+\frac{2a}{b}z+1=0 to see the location of the poles and if there are inside the unit circle or not. These are\begin{align*} -\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac} & =-\frac{\frac{2a}{b}}{2}\pm \frac{1}{2}\sqrt{\left ( \frac{2a}{b}\right ) ^{2}-4}\\ & =-\frac{a}{b}\pm \frac{1}{2}\sqrt{4\frac{a^{2}}{b^{2}}-4}\\ & =-\frac{a}{b}\pm \sqrt{\frac{a^{2}}{b^{2}}-1} \end{align*}
Since a>\left \vert b\right \vert then \frac{a^{2}}{b^{2}}>1 and the value under the square root is real. Hence both roots are real. Roots are \begin{align*} z_{1} & =-\frac{a}{b}+\sqrt{\frac{a^{2}}{b^{2}}-1}\\ z_{2} & =-\frac{a}{b}-\sqrt{\frac{a^{2}}{b^{2}}-1} \end{align*}
Now we need to decide the location of these poles. Let \frac{a}{b}=x. Where x>1 since a>\left \vert b\right \vert . Then the roots can be written as\begin{align*} z_{1} & =-x+\sqrt{x^{2}-1}\\ z_{2} & =-x-\sqrt{x^{2}-1} \end{align*}
Now \sqrt{x^{2}-1} is always smaller than x but \left ( \sqrt{x^{2}-1}-x\right ) can not be larger than 1 in magnitude. Hence z_{1} will always be inside the unit disk. On the other hand, \left ( \sqrt{x^{2}-1}+x\right ) will always be larger than 1 in magnitude (the sign is not important, we just wanted to know which pole is smaller or larger than 1 only. Therefore we conclude that z_{1} is inside the unit disk and z_{2} is outside.
Therefore, we need to find residue at z=0 and z=z_{1} and not at z=z_{2}. The function f\left ( z\right ) is from above is \begin{align*} f\left ( z\right ) & =\frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z^{2}+\frac{2a}{b}z+1\right ) }\\ & =\frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) } \end{align*}
Residue of f\left ( z\right ) at z=0
Since this pole is of order n=2, then\begin{align*} \operatorname{Residue} & =\lim _{z\rightarrow 0}\frac{d^{n-1}}{dz^{n-1}}\left ( \frac{\left ( z-z_{0}\right ) ^{n}f\left ( z\right ) }{\left ( n-1\right ) !}\right ) \\ & =\lim _{z\rightarrow 0}\frac{d}{dz}\left ( \left ( z-z_{0}\right ) ^{2}\frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z^{2}+\frac{2a}{b}z+1\right ) }\right ) \\ & =\lim _{z\rightarrow 0}\frac{d}{dz}\left ( z^{2}\frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z^{2}+\frac{2a}{b}z+1\right ) }\right ) \\ & =\lim _{z\rightarrow 0}\frac{d}{dz}\left ( \frac{2z^{2}-z^{4}-1}{z^{2}+\frac{2a}{b}z+1}\right ) \\ & =\lim _{z\rightarrow 0}\frac{\left ( 4z-4z^{3}\right ) \left ( z^{2}+\frac{2a}{b}z+1\right ) -\left ( 2z^{2}-z^{4}-1\right ) \left ( 2z+\frac{2a}{b}\right ) }{\left ( z^{2}+\frac{2a}{b}z+1\right ) ^{2}}\\ & =-\left ( -1\right ) \left ( \frac{2a}{b}\right ) \\ & =\frac{2a}{b} \end{align*}
Residue at z_{1}=-\frac{a}{b}+\sqrt{\frac{a^{2}}{b^{2}}-1}
Since this pole is of order 1, then the reside is\begin{align*} \operatorname{Residue} & =\lim _{z\rightarrow z_{1}}\left ( \left ( z-z_{1}\right ) f\left ( z\right ) \right ) \\ & =\lim _{z\rightarrow z_{1}}\left ( \left ( z-z_{1}\right ) \frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) }\right ) \\ & =\lim _{z\rightarrow z_{1}}\left ( \frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{z-z_{2}}\right ) \\ & =\frac{1}{z_{1}^{2}}\frac{2z_{1}^{2}-z_{1}^{4}-1}{z_{1}-z_{2}}\\ & =\frac{-1}{z_{1}^{2}}\frac{z_{1}^{4}-2z_{1}^{2}+1}{z_{1}-z_{2}}\\ & =\frac{-1}{z_{1}^{2}}\frac{\left ( z_{1}^{2}-1\right ) ^{2}}{z_{1}-z_{2}} \end{align*}
Let \frac{a}{b}=x, hence\sqrt{\frac{a^{2}}{b^{2}}-1}=\sqrt{x^{2}-1}. Therefore we can write z_{1}=-x+\sqrt{x^{2}-1} and z_{2}=-x-\sqrt{x^{2}-1} and now the above becomes\begin{align*} \operatorname{Residue} & =\frac{-1}{\left ( -x+\sqrt{x^{2}-1}\right ) ^{2}}\frac{\left ( \left ( -x+\sqrt{x^{2}-1}\right ) ^{2}-1\right ) ^{2}}{\left ( -x+\sqrt{x^{2}-1}\right ) -\left ( -x-\sqrt{x^{2}-1}\right ) }\\ & =\frac{-1}{2}\frac{\left ( \left ( -x+\sqrt{x^{2}-1}\right ) ^{2}-1\right ) ^{2}}{\left ( -x+\sqrt{x^{2}-1}\right ) ^{2}\sqrt{x^{2}-1}} \end{align*}
But \left ( -x+\sqrt{x^{2}-1}\right ) ^{2}=x^{2}+\left ( x^{2}-1\right ) -2x\sqrt{x^{2}-1}=2x^{2}-2x\sqrt{x^{2}-1}-1 and the above becomes\begin{align*} \operatorname{Residue} & =\frac{-1}{2}\frac{\left ( 2x^{2}-2x\sqrt{x^{2}-1}-1-1\right ) ^{2}}{\left ( 2x^{2}-2x\sqrt{x^{2}-1}-1\right ) \sqrt{x^{2}-1}}\\ & =\frac{-1}{2}\frac{\left ( 2x^{2}-2x\sqrt{x^{2}-1}-2\right ) ^{2}}{\left ( 2x^{2}-2x\sqrt{x^{2}-1}-1\right ) \sqrt{x^{2}-1}}\\ & =\frac{-1}{2}\frac{4\left ( x^{2}-x\sqrt{x^{2}-1}-1\right ) ^{2}}{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) \sqrt{x^{2}-1}}\\ & =-2\frac{\left ( \left ( x^{2}-1\right ) -x\sqrt{x^{2}-1}\right ) ^{2}}{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) \sqrt{x^{2}-1}} \end{align*}
Expanding gives\begin{align*} \operatorname{Residue} & =-2\frac{\left ( x^{2}-1\right ) ^{2}+\left ( x\sqrt{x^{2}-1}\right ) ^{2}-2\left ( x^{2}-1\right ) x\sqrt{x^{2}-1}}{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) \sqrt{x^{2}-1}}\\ & =-2\frac{\left ( x^{2}-1\right ) ^{2}+x^{2}\left ( x^{2}-1\right ) -2\left ( x^{2}-1\right ) x\sqrt{x^{2}-1}}{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) \sqrt{x^{2}-1}}\\ & =-2\frac{\left ( x^{2}-1\right ) \left ( x^{2}-1+x^{2}-2x\sqrt{x^{2}-1}\right ) }{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) \sqrt{x^{2}-1}}\\ & =-2\frac{\left ( x^{2}-1\right ) \left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) }{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) \sqrt{x^{2}-1}} \end{align*}
Dividing numerator and denominator by \left ( x^{2}-1\right ) \begin{align*} \operatorname{Residue} & =-2\frac{\sqrt{x^{2}-1}\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) }{\left ( 2x^{2}-1-2x\sqrt{x^{2}-1}\right ) }\\ & =-2\sqrt{x^{2}-1} \end{align*}
Since x=\frac{a}{b} then the above becomes \operatorname{Residue}=-2\sqrt{\frac{a^{2}}{b^{2}}-1} We found all residues. The sum is \sum \operatorname{Residue}=\frac{2a}{b}-2\sqrt{\frac{a^{2}}{b^{2}}-1} From the above we see now that\begin{align*} I & =\frac{1}{2bi}{\displaystyle \oint \limits _{C}} \frac{1}{z^{2}}\frac{2z^{2}-z^{4}-1}{\left ( z^{2}+\frac{2a}{b}z+1\right ) }dz\\ & =\frac{1}{2bi}\left ( 2\pi i\sum \operatorname{Residue}\right ) \\ & =\frac{1}{2bi}\left ( 2\pi i\left ( \frac{2a}{b}-2\sqrt{\frac{a^{2}}{b^{2}}-1}\right ) \right ) \\ & =\frac{\pi }{b}\left ( \frac{2a}{b}-2\sqrt{\frac{a^{2}}{b^{2}}-1}\right ) \\ & =\frac{\pi }{b}\left ( \frac{2a}{b}-\frac{2}{b}\sqrt{a^{2}-b^{2}}\right ) \\ & =\frac{2\pi }{b^{2}}\left ( a-\sqrt{a^{2}-b^{2}}\right ) \end{align*}
Hence the final result is \int _{0}^{2\pi }\frac{\sin ^{2}\theta }{a+b\cos \theta }d\theta =\frac{2\pi }{b^{2}}\left ( a-\sqrt{a^{2}-b^{2}}\right )