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 \[ \int _{0}^{2\pi }\frac{\sin ^{2}\theta }{a+b\cos \theta }d\theta =\frac{2\pi }{b^{2}}\left ( a-\sqrt{a^{2}-b^{2}}\right ) \]