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hadroneater
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Homework Statement
1. Evaluate
\int_{c_{2}(0)} f(z)dz = \int_{c_{2}(0)} \frac{z^{m}}{1+z^{3}}dz
Where c_{2}(0) is the circle of radius 2 centered at the origin with positive orientation (ccw).
I have done the question myself and compared it with the solution. However, I don't think I am understanding the use of the Laurent Series expansion right.
Homework Equations
The Attempt at a Solution
1. If m >= 0, then the integrand has 3 simple poles at z = -1, e^i(pi/3) and e^-i(pi/3).
If m < 0, then the integrand has the above poles and poles of order |m| at z = 0. The circular contour contains all such poles.
From here, what I did was simply apply the Residue Theorem.
For m >= 0
\int_{c_{2}(0)} \frac{z^{m}}{1+z^{3}}dz = 2i\pi(Res(-1) + Res(e^{\frac{i\pi}{3}}) + Res(e^{\frac{-i\pi}{3}}))
For m < = 0
\int_{c_{2}(0)} \frac{z^{m}}{1+z^{3}}dz = 2i\pi(Res(-1) + Res(e^{\frac{i\pi}{3}}) + Res(e^{\frac{-i\pi}{3}}) + Res(0))
The first 3 residues are easy to find because they are simple poles. The mth pole at zero is harder but still okay.
However, the solution does it differently. The first write the integrand as a Laurent Series about z = 0.
f(z) = \sum_{n = 0}^{\infty}z^{m}(-z^{3})^{n} = \sum_{n = 0}^{\infty}(-1)^{n}z^{m+3n}
Then for 1/z to appear, m + 3n = -1 hence m = -1 - 3n. Thus:
Res(0) = (-1)^{n} when m = -1 - 3n, n≥ 0
I can see why they would do that so that the residue at z = 0 would be easier to find. However, why is the summation for the Laurent Series from n = 0 and not n = -∞? Technically, isn't this just a power series?
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