Evaluate the integrals∮_C (f(z))dz over a contour C, where C is the boundary of a square with diagonal opposit

Evaluate the integrals∮_C (f(z))dz over a contour C, where C is the boundary of a square with diagonal opposite corners at z = −(1 + i )R and z = (1 + i )R, where R > a > 0, and where f (z) is given by the following:
1)e^z/(z-(πi/4)a)^2

2)sin z/(z^2)

1) The integrand has singularity at z = πai/4, which is a double pole.
This is inside C, because Im(πai/4) = πa/4 < πR/4 < R.

So, Cauchy's Integral Theorem yields
∫c e^z dz/(z - πai/4)^2 = 2πi * (d/dz) e^z {at z = 0} = 2πi.
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2) Note that the only singularity of sin z/z^2 is at z = 0.
This is a pole of order 1, because sin z/z = 1 - z^2/3! + ... is analytic at z = 0.
(we can thus define sin(z)/z {at z = 0} as 1 - 0 = 1.)

By Cauchy's Integral Theorem,
∫c sin z dz/z^2 = ∫c (sin(z)/z) dz / (z - 0) = 2πi * sin(z)/z {at z = 0} = 2πi * 1 = 2πi.
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I hope this helps!