Complex Analysis Residues at Poles

[beefcake24]

Homework Statement

Find the residue at each pole of zsin(pi*z)/(4z^2 - 1)

Homework Equations

An isolated singular point z0 of f is a pole of order m if and only if f(z) can be written in the form:

f(z) = phi(z)/(z-z0)^m

where phi(z) is analytic and nonzero at z0. Moreover,

Res(z=z0) f(z) = phi(z0) if m = 1

and

Res(z=z0) f(z) = phi^(m-1)(z0)/(m-1)! if m >= 2

The Attempt at a Solution

I don't know what I'm missing here, the problem seems really easy. I factored it to

z*sin(pi*z)/[(2z+1)(2z-1)]

so f(z) has simple poles at z = 1/2 and z = -1/2

For z = 1/2, we have f(z) = phi(z)/(z-1/2) where phi(z) = z*sin(pi*z)/(z+1/2)

Plugging in z = 1/2 in phi(z) I get a residue of 1/4.

Similarly, I get a residue of -1/4 at the pole of z = -1/2.

But the answer is -1/8 and 1/8 for the residues respectively, and I can't figure out what I'm doing wrong.

 

[micromass]

Homework Statement

Find the residue at each pole of zsin(pi*z)/(4z^2 - 1)

Homework Equations

An isolated singular point z0 of f is a pole of order m if and only if f(z) can be written in the form:

f(z) = phi(z)/(z-z0)^m

where phi(z) is analytic and nonzero at z0. Moreover,

Res(z=z0) f(z) = phi(z0) if m = 1

and

Res(z=z0) f(z) = phi^(m-1)(z0)/(m-1)! if m >= 2

The Attempt at a Solution

I don't know what I'm missing here, the problem seems really easy. I factored it to

z*sin(pi*z)/[(2z+1)(2z-1)]

so f(z) has simple poles at z = 1/2 and z = -1/2

For z = 1/2, we have f(z) = phi(z)/(z-1/2) where phi(z) = z*sin(pi*z)/(z+1/2)

I do not see where your phi comes from. You seem to have done

$$(2z+1)(2z-1)=(z+1/2)(z-1/2)$$

which is not true. You are missing a factor 4 there.

Plugging in z = 1/2 in phi(z) I get a residue of 1/4.

Similarly, I get a residue of -1/4 at the pole of z = -1/2.

But the answer is -1/8 and 1/8 for the residues respectively, and I can't figure out what I'm doing wrong.

 

[beefcake24]

Ohh haha that was stupid. Thanks!