Compute the closed contour integral of 1/sinh[2z] over the contour |z|=2
This integral is the sum of the residues of the poles of 1/sinh[2z] that lie inside the contour times 2 pi i.
The poles of 1/sinh[2z] are (k+1/2) i pi, k an integer. Only 2 of these lie inside |z|=2, +/- i pi/2.
The residue of 1/sinh[2z] at +/- i pi/2 is -1/2 for each and thus their sum is -1.
Thus the integral should be 2 pi i * -1 = -2 pi i
My book says the answer is -pi i
Can anyone see where the factor of 1/2 comes from?
This integral is the sum of the residues of the poles of 1/sinh[2z] that lie inside the contour times 2 pi i.
The poles of 1/sinh[2z] are (k+1/2) i pi, k an integer. Only 2 of these lie inside |z|=2, +/- i pi/2.
The residue of 1/sinh[2z] at +/- i pi/2 is -1/2 for each and thus their sum is -1.
Thus the integral should be 2 pi i * -1 = -2 pi i
My book says the answer is -pi i
Can anyone see where the factor of 1/2 comes from?