Question

Let C be any simple closed contour, described in the positive sense in the z plane and write g(z)= Integral over C of (s^3+2s)/(s−z)^3 and show that g(z)=6*pi*i*z when z is inside C and g(z)=0 when z is outside

Answer 1

First, if z is outside of C, then the function f(s) = (s^3+2s)/(s−z)^3 is analytic on the interior of C, so since C is a closed curve, one of the major theorems (Cauchy I think?) tells us that the integral is zero, so g(z) = 0.

Next, if z is inside of C, then the function f(s) = (s^3+2s)/(s−z)^3 has a pole of order 3 contained in the interior of C. You should be able to apply Cauchy's Integral Formula then: let h(s) = s^3+2s. Then h''(s) = 6s, so
\[ 6z = \frac{2!}{2\pi i}\int \frac{h(z)}{(s-z)^3}ds\]
and shuffling terms around gets
\[ 6z \pi i = g(z).\]