Author Topic: Contour integration (Read 1343 times) Tweet Share

QuantumJG · « **on:** April 20, 2011, 06:47:50 pm »

In a lecture today we evaluated a integral:

$\oint_{\Gamma} \dfrac{3z - 2}{z^2 - z} dz$

Where,

$\Gamma = \{ |z| + |z-1| = 3 \}$

Our lecturer evaluated it to be 6πi

I sort of understood how he did it, but he really rushed through his steps.

/0 · « **Reply #1 on:** April 20, 2011, 08:09:19 pm »

You use Cauchy's residue theorem

$\oint_{\gamma} f(z)\,dz=2\pi i\sum_{poles} \mbox{Res}_{z=one\ of\ the\ poles\ inside\ the\ contour} f(z)$

$f(z)=\frac{3z-2}{z^2-z}$ has poles where it is not analytic, i.e. where $z^2-z=0\Rightarrow z=0,1$ . Both are poles of order 1, i.e. none of the poles are repeated, so the formula for the residues is $\lim_{z \to a}(z-a)f(z)$ , $a=0,1$ .

If we look at the contour, both $z=0,1$ are inside the contour, being foci of the ellipse, so we need to find the residues at each point.

We have:

$\mbox{Res}_{z=0} f(z)=\lim_{z\to 0}(z-0)f(z)=\lim_{z\to 0}z\frac{3z-2}{z^2-z}=\lim_{z \to 0}\frac{3z-2}{z-1}=2$

$\mbox{Res}_{z=1} f(z)=\lim_{z \to 1}(z-1)f(z)=\lim_{z \to 1} (z-1)\frac{3z-2}{z^2-z}=\lim_{z\to 1}\frac{3z-2}{z}=1$

Hence, we have

$Integral = 2\pi i \left(1+2\right)=6\pi i$

Is that kinda what u wanted, or did u need help understanding the origins of the residue formula/cauchy's residue theorem?

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Author Topic: Contour integration (Read 1343 times) Tweet Share

QuantumJG

Contour integration

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Re: Contour integration

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