# Integration along circular path

1. Feb 10, 2006

### Benny

Hi, I'm having trouble with two integrals.

$$\int\limits_{\left| z \right| = 1}^{} {\frac{{e^z }}{{z^n }}dz}$$

$$\int\limits_{\left| z \right| = 1}^{} {\frac{{\cosh z}}{{z^3 }}} dz$$

For first one I wrote

$$\frac{{e^z }}{{z^n }} = \frac{{e^z }}{{\left( {z - 0} \right)z^n }} = \frac{{f\left( z \right)}}{{\left( {z - 0} \right)}}$$.

The main theorem which was covered in the relevant theory section was Cauchy's integral formula. By inspection f(z) is not even differentiable at z = 0 because it isn't continuous there? So can I even apply the integral formula?

Ignoring that, if n <=0 then f(z) = exp(z)z^(1-n) = 0 for z = 0. So the integral is zero for n <=0. But what about for n > 0? There's a factorial in the answer. I don't really know what to do because ofr n > 0 then f(z) = exp(z)/z^(positive power) so if I plug in z = 0 I get an undefined answer.

Last edited: Feb 11, 2006
2. Feb 11, 2006

### Tide

HINT: Do a Taylor series expansion of the integrand about z = 0 and the integral will be the coefficient of the 1/z term (within a factor of [itex]2\pi i[/tex]).

3. Feb 11, 2006

### Benny

Ok I'll try that, thanks for the help. I made a slight error in my working for the first one. it should be (exp(z))/(z^n) = (exp(z))/((z-0)z^(n-1)) = f(z)/(z-0).