Using Cauchy Integral, solve: $\oint_C \frac{1}{(2z+1)^2}dz$ without use residue theorem

Question

C is a unitary circle.

If I simplify, I get: $\oint_C \frac{1/2}{(z+1/2)(z+1/2)}$ How can i solve? If I use Cauchy Integral Theorem, $z_0 = -1/2$ and f(z) = 2, so answer is $2\pi i * (-1/2)$?

Idk how to solve with 1/(z-a)^n

Answer 1

Unlike the logarithm, $-\tfrac{1/2}{2z+1}$ doesn't have branches, so an integral of $\tfrac{1}{(2z+1)^2}$ around a closed loop is $0$.