Residues

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Enter homework on residues here.

++Question 1

Compute the residue at the poles of each of the following functions.

a) $f(z)=\frac{1}{z^{2}-6z+8}$

(1)
\begin{align} \frac{1}{z^{2}-6z+8} \end{align}
(2)
\begin{align} \frac{1}{(z-4)(z-2)} \end{align}

Simple poles at $z=4$ and $z=2$.

(3)
\begin{align} Res(4)= \frac{1}{2z-6} = \frac{1}{2(4)-6} = \frac{1}{2} \end{align}
(4)
\begin{align} Res(2)= \frac{1}{2z-6} = \frac{1}{2(2)-6} = -\frac{1}{2} \end{align}

b) $f(z)=\frac{1}{z^{3}+1}$

Simple poles at z=-1 ,$z=\frac{1+i\sqrt{3}}{2}$ , and z=$z=\frac{1+i\sqrt{3}}{2}$

(5)
\begin{align} Res(-1)= \frac{1}{3z^2} = \frac{1}{3} \end{align}
(6)
\begin{align} Res(\frac{1+i\sqrt{3}}{2})= (\frac{1}{{3}(\frac{1+i\sqrt{3}}{2})^2})= (\frac{1}{{3}(\frac{-1+i\sqrt{3}}{2})})= (\frac{1}{(\frac{-3+3i\sqrt{3}}{2})}) \\= \frac{2}{{-3+3i\sqrt{3}}}(\frac{-3-3i\sqrt{3}}{-3-3i\sqrt{3}})=\frac{2(-3-3i\sqrt{3})}{36}=\frac{(-3-3i\sqrt{3})}{18} \\=\frac{(-1-i\sqrt{3})}{6} \\\\Res(\frac{1-i\sqrt{3}}{2})=(\frac{1}{{3}(\frac{1-i\sqrt{3}}{2})^2})=(\frac{1}{{3}(\frac{-1-i\sqrt{3}}{2})})=(\frac{1}{(\frac{-3-3i\sqrt{3}}{2})}) \\= \frac{2}{{-3-3i\sqrt{3}}}(\frac{-3+3i\sqrt{3}}{-3+3i\sqrt{3}})=\frac{-6+ 6i\sqrt{3}}{36} \\=\frac{(-1+i\sqrt{3})}{6} \end{align}
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