$ \oint \frac {3z^3 + 2}{(z-1)(z^2+9)} dz$ taken counterclockwise around circles:
(a) |z-2| = 2; (b) |z| = 4
My Attempt:
The circle of radius 2 centered at z = 2 only encloses the singularity z = 1 but I'm not sure whether I can use the Cauchy integral formula with it.
The circle of radius 4 centered at the origin encloses all the singular points, namely:
z = 1, +3i, -3i (from $(z^2 +9) -> (z-3i)(z+3i) $)
So can I just do
$ \oint \frac {3z^3 + 2}{(z-1)(z^2+9)} dz = 2\pi i (\sum residues) ?$
Best Answer
First of all, think about the contours and what singularities they enclose. You should know that by Cauchy's Theorem a closed integral around a function that is analytic on an in the contour is zero. However, we are going to have some singularities here because of the denominator.
You did this. Nice!
Use Cauchy's Integral Formula.
Part (a) is a circle of radius two centered at 2. It will enclose z = 1.
$\oint \frac{3z^3+2}{(z-1)(z+3i)(z-3i)}dz = 2\pi i \left[\left.\left(\frac{3z^3+2}{z^2+9}\right)\right|_{z=1}\right]$
Part (b) will enclose everything.
$\oint \frac{3z^3+2}{(z-1)(z+3i)(z-3i)}dz = 2\pi i \left[\left.\left(\frac{3z^3+2}{z^2+9}\right)\right|_{z=1}+\left.\left(\frac{3z^3+2}{(z-1)(z+3i)}\right)\right|_{z=3i}+\left.\left(\frac{3z^3+2}{(z-1)(z-3i)}\right)\right|_{z=-3i}\right]$