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QUESTION

If $C$ is the unit circle centred on the origin, evaluate by the
residue theorem



$$\textrm{(i)}\int_C\frac{\cos z}{2z^2-7z}\,dz\ \ \textrm{(ii)}\ \
\int_C\frac{2z+1}{2z^4-z^3}\,dz\ \ \textrm{(iii)}\ \
\int_C\frac{e^z}{8z^3-1}\,dz$$



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ANSWER
\begin{description}

\item[(i)]
 $\ds\int_{|z|=1}\frac{\cos z}{2z62-7z}\,dz$ has simple poles at
 $z=0$
 and $z=\frac{7}{2}$. Hence
 $$\int_{|z|=1}\frac{\cos z}{2z^2-7z}\,dz=2 \pi i \textrm{Res} (0)=2 \pi i \lim_{z \to 0}
 \frac{\cos z}{4z-7}=-\frac{2 \pi i}{7}.$$

\item[(ii)]
 $\ds\int_{|z|=1}\frac{2z+1}{2z^4-z^3}\,dz$

 has a simple pole at $z=\frac{1}{2}$ and a triple pole at $z=0$.
 $$\textrm{Res}\left(\frac{1}{2}\right)=\lim_{z \to
 \frac{1}{2}}\frac{2z+1}{8z^3-3z}=8$$
 $$\textrm{Res}(0)=\frac{1}{2!}\lim_{z \to
 \frac{1}{2}}\left(\frac{d}{dz}\right)^2\frac{2z+1}{2z-1}=\ldots$$
 Easier:
 \begin{eqnarray*}
 \frac{1}{z^3}\frac{2z+1}{2z-1}&=&-\frac{1}{z^3}(2z+1)\frac{1}{1-2z}\textrm{(use
 geometric series)}\\
 &=&-\frac{1}{z^2}(1+2z)(1+2z+4z^2+\ldots)\\
 &=&-\frac{1}{z^3}(1+4z+8z^2+\ldots)\\
 &\Rightarrow & \textrm{Res}(0)=-8 \Rightarrow I=0
 \end{eqnarray*}

\item[(iii)]
 $\ds I=\int_{z=0}\frac{e^z}{8^3-1}\,dz$ has simple poles at
  $z=\frac{1}{2},\ \frac{1}{2}e^\frac{2 \pi i}{3},\
 \frac{1}{2}e^\frac{4 \pi i}{3}$


 $$\textrm{Res}\left(\frac{e^z}{8z^3-1},z_0\right)=\lim_{z \to z_0}
 \frac{e^z}{24z^2}=\frac{e^{z_0}}{24z_0}=\frac{z_0e^{z_0}}{24z_0^3}
 =\frac{z_0e^{z_0}}{3}$$


 \begin{eqnarray*}
 I&=&2 \pi i
 \frac{1}{3}\left(\frac{1}{2}e^\frac{1}{2}+\frac{1}{2}e^\frac{2 \pi
 i}{3}e^{-\frac{1}{4}+\frac{\sqrt{3}}{4}i}+\frac{1}{2}e^\frac{4
 \pi i}{3}e^{-\frac{1}{4}-\frac{\sqrt{3}}{4}i}\right)\\
 &=&\frac{\pi i}{3}\left[e^\frac{1}{2}+e^{-\frac{1}{4}}2 \cos \left(\frac{2
 \pi}{3}+\frac{\sqrt{3}}{4}\right)\right]
 \end{eqnarray*}

\end{description}

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