Evaluating Integral on Circular Contour C: Quick Question on Residues

[kreil]

Homework Statement

Evaluate the following integral, given that C is the circular contour of radius greater than 2 centered at the origin.

$$I_a=\int_C\frac{z^2-1}{(2z-1)(z^2-4)^2}dz$$

The Attempt at a Solution

I has a simple pole at z=1/2 and two double poles, at z=2 and z=-2...all of which are enclosed by the contour C. By the residue theorem,

$$I_a=2 \pi i \left[ Res[f(1/2)]+Res[f(2)]+Res[f(-2)] \right]$$

My problem is just that the answers I am getting for the residues are strange...

$$Res[f(1/2)]=\lim_{z \rightarrow 1/2} (2z-1) \frac{z^2-1}{(2z-1)(z^2-4)^2}=\lim_{z \rightarrow 1/2} \frac{z^2-1}{(z^2-4)^2}= \frac{-3/4}{225/16}=-\frac{12}{225}$$

Is this answer correct? When I compute the residue for z=2 I get an even weirder answer (10/192) and these strange answers are making me question whether what I'm doing is correct (note that I used the multipole formula for z=2)

 

[Dick]

It's not quite correct. You need to multiply by (z-1/2), not (2z-1). But I don't see anything weird about those answers. I didn't get the same thing you did for z=2, though.

 

[kreil]

It's not quite correct. You need to multiply by (z-1/2), not (2z-1). But I don't see anything weird about those answers. I didn't get the same thing you did for z=2, though.

Hm, ok. So for the 1/2 residue the answer is -6/225, correct?

I found a mistake in the z=2, thanks.

 

[Dick]

Hm, ok. So for the 1/2 residue the answer is -6/225, correct?

I found a mistake in the z=2, thanks.

That's what I get. Or -2/75.