Mark
Compute
\int_{\gamma} \frac{z}{(z+1)(z-2)(z-4)(z-8)}dz,
where \gamma(t) = 3+2e^{2\pi it} for 0\leq t \leq 1.
An archive the questions from Mark's Fall 2018 Complex Variables Course.
An integral
warsh
\int_{\gamma} g(z)dz = \int_{\gamma} \frac{z}{(z+1)(z-2)(z-4)(z-8)}dz
\gamma(t) = 3+2e^{2\pi it} parametrizes a path centered at 3 with radius 2. The poles of g(z) are at z=-1, 2, 4, 8. The only poles at lie inside of the path of integration are z=2 and z=4. We must break \gamma into two paths which only have one pole inside, so that we can use Cauchy’s Integral Formula:
f(w)=\frac{1}{2\pi i} \int_{\gamma}\frac{f(z)}{z-w}dz.
\begin{align}
\int_{\gamma} \frac{z}{(z+1)(z-2)(z-4)(z-8)}dz &= \int_{\gamma_{1}} \frac{z/((z+1)(z-4)(z-8))}{z-2}dz \\
& +\int_{\gamma_{2}} \frac{z/((z+1)(z-2)(z-8))}{z-4}dz,
\end{align}
where \gamma_{1} is a positively oriented path with only the pole z=2 inside and \gamma_{2} is a positively oriented path with only the pole z=4 inside.
We can now evaluate each of this integrals using Cauchy’s Integral Formula.
\begin{align}
2\pi if(w)= \int_{\gamma}\frac{f(z)}{z-w}dz &= \int_{\gamma_{1}} \frac{z/((z+1)(z-4)(z-8))}{z-2}dz \\
&= 2 \pi i \frac{2}{(2+1)(2-4)(2-8)} = \frac{4\pi i}{36} = \frac{\pi i}{9},
\end{align}
2\pi if(w)= \int_{\gamma}\frac{f(z)}{z-w}dz = \int_{\gamma_{1}} \frac{z/((z+1)(z-2)(z-8))}{z-4}dz = 2 \pi i \frac{4}{(4+1)(4-2)(4-8)} = \frac{8\pi i}{-40} = -\frac{\pi i}{5}
\int_{\gamma_{1}} \frac{z/((z+1)(z-4)(z-8))}{z-2}dz+\int_{\gamma_{2}} \frac{z/((z+1)(z-2)(z-8))}{z-4}dz= \frac{\pi i}{9} -\frac{\pi i}{5} = -\frac{4 \pi i}{45}
So \int_{\gamma} g(z)dz = \int_{\gamma} \frac{z}{(z+1)(z-2)(z-4)(z-8)}dz = -\frac{4 \pi i}{45}, where \gamma is a circle of radius 2 centered at 3.