I'm asked to integrate the function $f(z) = \dfrac{1}{z-\frac{1}{2}}$ over the circle centered at the origin with radius $3$ with positive orientation. I solved the integral with Cauchy's Integral formula and I get that $\displaystyle \int_\gamma\dfrac{1}{z-\frac{1}{2}}\mathrm{d}z$ is $2\pi i$ and then since I'm just starting with these integrals I wanted to check using a parametrization, mainly $\gamma(t) = 3 e^{it} $ for $ 0\leq t \leq 2\pi $ and I end up with this integral $\displaystyle \int_0^{2\pi} \frac{3ie^{it}}{3e^{it}-\frac{1}{2}}\mathrm{d}t$ trying different functions I found that the integral could be solved this way:
$$\begin{aligned} \int_0^{2\pi} \left(Log\left(3e^{it}-\frac{1}{2}\right)\right)^{\prime} \mathrm{d}t &= Log\left(3e^{it}-\frac{1}{2}\right) \biggr|_0^{2\pi}\\ &= Log\left(3e^{i2\pi}-\frac{1}{2}\right) - Log\left(3e^{i0}-\frac{1}{2}\right) \\ &= Log\left(3-\frac{1}{2}\right) - Log\left(3-\frac{1}{2}\right) \\ & =0\end{aligned}$$
Can someone explain me this inconsistency please? And also I'm not sure what log I'm supposed to use, the real valued $\ln$ or well the one I assume $Log$.
Taking the branch of the logarithm $f(z)=\log (z-\frac{1}{2})$:
$$ 0 \le \arg \left(z-\frac{1}{2} \right)< 2\pi,$$ with a branch point at $z_0=1/2$ and a cut from the point $z_0$ along the positive real axis to the point at infinity,
$$\oint \frac{dz}{z-\frac{1}{2}}= \lim_{\theta\uparrow 2\pi} \log\left(3e^{i\theta} - \frac{1}{2}\right)- \log\left(3e^{i\theta}-\frac{1}{2}\right)_{\theta=0} = 2\pi i - 0 = 2\pi i.$$
The function $f(z)$ is holomorphic along the path (but not at the end points).