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PostPosted: Wed May 02, 2018 9:37 pm 

Joined: Wed Nov 15, 2017 12:37 pm
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Can someone help me with this one? I want to compute this integral without using the residue theorem. How is it solved if one uses Cauchy's integral theorem?
Assume that $f(z)=\frac{1}{(z-2)^2(z-4)}$, which has singularities at $2$ and $4$ and suppose we have to compute \[\displaystyle\oint_{C}{f(z)\,dz}\] with $C$ been the positive oriented circle $|z|=3$, or $|z|=5$ (contour containing one or both singular points).


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PostPosted: Thu May 03, 2018 6:23 am 
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The function $f(z)=\frac{1}{(z-2)^2(z-4)}$ is defined and is holomorphic on $\mathbb{C}\setminus\{2,4\}$. The disk $D_1=\big\{{z\in\mathbb{C}\;|\;|z|\leqslant3}\big\}$ containing the second order pole $z_1=2$, but not the simple pole $z_2=4$. By Cauchy's integral formula we have \begin{align*}
\displaystyle\oint_{|z|=3}\frac{1}{(z-2)^2(z-4)}\,dz&=\oint_{|z|=3}{\dfrac{\frac{1}{z-4}}{(z-2)^2}\,dz}\\
&=\frac{2\pi i}{1!}\,\frac{d}{dz}\Big(\frac{1}{z-4}\Big)\bigg|_{z=2}\\
&=2\pi i\,\Big(-\frac{1}{4}\Big)\\
&=-\frac{\pi i}{2}\,.
\end{align*}
For the second integral, see that $f(z)=\frac{1}{(z-2)^2(z-4)}=\frac{1}{4(z-4)}-\frac{z}{4(z-2)^2}$. So,
\begin{align*}
\displaystyle\oint_{|z|=5}\frac{1}{(z-2)^2(z-4)}\,dz&=\frac{1}{4}\oint_{|z|=5}{\frac{1}{z-4}\,dz}-\frac{1}{4}\oint_{|z|=5}\frac{z}{(z-2)^2}\,dz\\
&=\frac{2\pi i}{4}-\frac{1}{4}\frac{2\pi i}{1!}\,\frac{d}{dz}z\,\bigg|_{z=2}\\
&=\frac{\pi i}{2}-\frac{\pi i}{2}\\
&=0\,.
\end{align*}

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