Use partial fractions to evaluate the following integral ∮_C (f(z))dz

Use partial fractions to evaluate the following integral ∮_C (f(z))dz where
C is the unit circle centered at the origin, and f (z) is given by the following:f(z)1/(z(z + 1/2)(z-2))

Note that of the singularities of f (namely z = 0, -1/2, 2), only z = 2 is not in C.

So, we can rewrite the integral as
∫c dz/(z(z + 1/2)(z - 2))
= ∫c (1/(z-2)) dz / (z(z + 1/2))
= ∫c (2/(z-2)) [1/z - 1/(z + 1/2)], by partial fractions
= ∫c (2/(z-2)) dz/(z - 0) - ∫c (2/(z-2)) dz/(z + 1/2)
= 2πi * 2/(z - 2) {at z = 0} - 2πi * 2/(z - 2) {at z = -1/2}, by Cauchy's Integral Theorem
= -2πi/5.

I hope this helps!