Computing ∫|z|=2π (π sin(z))/(z(z-π)) dz with Cauchy's formula

[phyzmatix]

Homework Statement

Use Cauchy's Integral Formula to compute the following integral:

$$\int_{|z|=2\pi}\frac{\pi \sin(z)}{z(z-\pi)}dz$$

Homework Equations

$$\int_C \frac{f(z)dz}{z-z_0}=2\pi if(z_0)$$

The Attempt at a Solution

If we let

$$f(z)=\frac{\pi \sin(z)}{z}$$

and

$$z_0=\pi$$

Then

$$\int_{|z|=2\pi}\frac{\pi \sin(z)/z}{(z-\pi)}dz=0$$

Is this correct? I'm concerned about the fact that the function isn't differentiable at z=pi and that pi is an interior point of C...Part of me thinks that this integral cannot be evaluated with Cauchy's formula, but I do not yet understand the subject matter well enough to know for sure.

 

[Susanne217]

Homework Statement

Use Cauchy's Integral Formula to compute the following integral:

$$\int_{|z|=2\pi}\frac{\pi \sin(z)}{z(z-\pi)}dz$$

Homework Equations

$$\int_C \frac{f(z)dz}{z-z_0}=2\pi if(z_0)$$

The Attempt at a Solution

If we let

$$f(z)=\frac{\pi \sin(z)}{z}$$

and

$$z_0=\pi$$

Then

$$\int_{|z|=2\pi}\frac{\pi \sin(z)/z}{(z-\pi)}dz=0$$

Is this correct? I'm concerned about the fact that the function isn't differentiable at z=pi and that pi is an interior point of C...Part of me thinks that this integral cannot be evaluated with Cauchy's formula, but I do not yet understand the subject matter well enough to know for sure.

I think you need to look at where the integral has the socalled simple pole

and $$z= \pi$$ is a simple pole.

 

[phyzmatix]

Cheers Susanne! Going to try that angle