Re: Cauchy Integral Formula

... may be that the reason is that in $\displaystyle z=\frac{1}{3}$ 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

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Cauchy Integral Formula

Ahh, I just discovered the generalised CIF equation.

Still not sure how to get the correct solution though, I did:

$\displaystyle \frac{2\pi i}{!1} 2ie^{2iz}$

= $\displaystyle -4\pi e^{\frac{2i}{3}}$

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

Re: Cauchy Integral Formula

For a pole of order n is...

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

In our case is...

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

... so that...

$\displaystyle 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 $\displaystyle z=\frac{1}{3}$ is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Cauchy Integral Formula

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

Thanks.