Evaluate ∮_C (f(z))dz

Evaluate the integrals ∮_C (f(z))dz over a contour C, where C is the boundary of a square withdiagonal 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
a)(sin z)/z^2
b)coshz/z

Thanks for your help

a) The only singularity of sin(z)/z^2 is at z = 0.

Since sin(z)/z has removable singularity at z = 0 (look at limit or Laurent series at z = 0),
z = 0 is a simple pole of sin(z)/z^2, which is inside C.

Hence, Cauchy's Integral Formula yields
∫c sin z dz/z^2
= ∫c (sin(z)/z) dz/(z - 0)
= 2πi * (sin(z)/z) {at z = 0}; remember this is a removable singularity
= 2πi * 1
= 2πi.
--------------
b) The only singularity of cosh(z)/z is at z = 0, which is a simple pole inside C.

Hence, Cauchy's Integral Formula yields
∫c cosh z dz/z
= 2πi * (cosh z) {at z = 0};
= 2πi * 1
= 2πi.

I hope this helps!