Compute $\int_{\gamma}\frac{\log(1+z)}{ (z-\frac{1}{2})^3}dz$ Compute
$$\int_{\gamma}\frac{\log(1+z)}{(z-\frac{1}{2})^3}dz$$
Where $\gamma:[0,2\pi]\to \Bbb{C}$ is given by $\gamma(t)=\frac{2}{3}e^{it}$.

Im quite confused in which branch of the logarithm should I use. And how should I proceed once I've chosen the correct one?
log(1+z) is analytic if we remove the ray (-inf,-1] and the given curve is in the resulting region. Thus from Cauchy formula f''(a)=2/2.pi.i.integral round C of f(z)/(z-a)^3 we get that the integral is (log(1+z))''.pi.i at z=1/2. It is i.pi.1/(9/4)
log(1+z) is analytic if we remove the ray (-inf,-1] and the given curve is in the resulting region. Thus from Cauchy formula f''(a)=2/2.pi.i.integral round C of f(z)/(z-a)^3 we get that the integral is (log(1+z))''.pi.i at z=1/2. It is i.pi.1/(9/4)