Cauchy Integral Formula

**Glitch** · June 16th 2011, 01:55 AM

The question
Evaluate the following integral using Cauchy Integral Formula,
$\oint_{|z| = 4}{\frac{e^{2iz} \ dz}{(3z - 1)^2}}$

My attempt
Not analytic at:
$(3z - 1)^2 = 0$
$z = \frac{1}{3}$

This is inside the contour, so we use CIF as follows:
$2\pi i e^{\frac{2i}{3}}$

However, the solution is $\frac{-4\pi}{9}e^{\frac{2i}{3}}$

Where have I gone wrong? Thanks.

**chisigma** · June 16th 2011, 02:23 AM

... may be that the reason is that in $z=\frac{1}{3}$ the function has a pole with multeplicity two... in that case the formula for the residue is different respect to the case of a pole with multeplicity one...

Kind regards

$\chi$ $\sigma$

**Glitch** · June 16th 2011, 03:00 AM

Ahh, I just discovered the generalised CIF equation.

Still not sure how to get the correct solution though, I did:
$\frac{2\pi i}{!1} 2ie^{2iz}$
= $-4\pi e^{\frac{2i}{3}}$

Not sure where the divide by 9 is coming from. :/

**chisigma** · June 16th 2011, 08:00 AM

For a pole of order n is...

$r= \lim_{z \rightarrow z_{0}} \frac{d^{n-1}}{d z^{n-1}} \{(z-z_{0})^{n}\ f(z) \}$ (1)

In our case is...

$f(z)= \frac{1}{9}\ \frac{e^{2 i z}}{(z-\frac{1}{3})^{2}}$ (2)

... so that...

$r= \lim_{z \rightarrow \frac{1}{3}} \frac{d}{d z} \frac{e^{2 i z}}{9} = \frac{2i}{9}\ e^{\frac{2}{3} i}$ (3)

... and for any closed path c containing the point $z=\frac{1}{3}$ is...

$\int_{c} f(z)\ dz = 2 \pi i r = - \frac{4 \pi}{9}\ e^{\frac{2}{3} i}$ (4)

Kind regards

$\chi$ $\sigma$

**Glitch** · June 16th 2011, 08:40 AM

Ahh, I just had to pull the 3 out of the denominator.

Thanks.

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