Cauchy's integral formula

• April 15th 2010, 01:01 AM
piglet
Cauchy's integral formula
$\int_{C}\frac{f(z)}{z-z0}dz = 2\pi.i.f(z0)$

Im using cauchy's integral formula, see above to solve the following integral

$\int_{C}\frac{z}{z^{2}-1}dz$ where $C$ is a circle of radius 4 centred at 0.

I have deduced that the function $\frac{z}{z^{2}-1}$ is not analytic at $z = +i$ $z = -i$

I let $f(z) = \frac{z}{z+i}$ and $z0 = -i$

so filling into cauchy's integral formula i got

$\int_{C}\frac{z}{(z+i)(z-i)}dz = \frac{f(z)}{z-i} = 2\pi.i.f(i)$

Answer = $\pi.i$

My problem lies in the fact that both $z = +i$ $z = -i$ both lie in $C$. Therefore do i have to take both singularites into account when evaluating the integral using cauchy's integral formula???

Thanks,
Piglet
• April 15th 2010, 02:40 AM
mr fantastic
Quote:

Originally Posted by piglet
$\int_{C}\frac{f(z)}{z-z0}dz = 2\pi.i.f(z0)$

Im using cauchy's integral formula, see above to solve the following integral

$\int_{C}\frac{z}{z^{2}-1}dz$ where $C$ is a circle of radius 4 centred at 0.

I have deduced that the function $\frac{z}{z^{2}-1}$ is not analytic at $z = +i$ $z = -i$

I let $f(z) = \frac{z}{z+i}$ and $z0 = -i$

so filling into cauchy's integral formula i got

$\int_{C}\frac{z}{(z+i)(z-i)}dz = \frac{f(z)}{z-i} = 2\pi.i.f(i)$

Answer = $\pi.i$

My problem lies in the fact that both $z = +i$ $z = -i$ both lie in $C$. Therefore do i have to take both singularites into account when evaluating the integral using cauchy's integral formula???

Thanks,
Piglet

Yes. See example 30.2 here: http://math.furman.edu/~dcs/courses/...lecture-30.pdf
• April 15th 2010, 01:44 PM
zzzoak
$
\frac{z}{z^{2}-1} \: = \:\frac{1}{2(z-1)} \: + \frac{1}{2(z+1)} \:

$