Q:

Show that $\oint_{\gamma} \frac{Log z}{z}dz=0$

I need to prove that 
$$\oint_{\gamma}\frac{\log z}{z}dz=0$$ 
where $\log z= \log |z|+ \text{Arg } z$ ($\text {Arg}$ is the principal argument), $\gamma:z=e^{it},\quad 0\leq t\leq 2\pi$.   
I know that I cannot solve it with Cauchy Formula, because $(\log z)'=\frac{1}{z}$, so $(\log 0)'$ does not exists. I have tried to use the  parameterization of $\gamma$, but I am not pretty sure how to deal with the limits of $\text{Arg } z$. 

A:

For branch on positive real axis:
$$\begin{align}
\oint\frac{\log z}{z}dz 
&=\int^{2\pi}_0\frac{\log|e^{it}|+i\arg{e^{it}}}{e^{it}}ie^{it}dt \\
&=i\int^{2\pi}_0 (\log 1+it)dt \\
&=-2\pi^2\ne0
\end{align}
$$

For principal branch:
$$\begin{align}
\oint\frac{\log z}{z}dz 
&=\int^{2\pi}_0\frac{\log|e^{it}|+i\arg{e^{it}}}{e^{it}}ie^{it}dt \\
&=i\int^{2\pi}_0 (\log 1+i\operatorname{Arg}(e^{it})dt \\
&=i\int^\pi_0 i\operatorname{Arg}(e^{it})dt +i\int^{2\pi}_{\pi} i\operatorname{Arg}(e^{it})dt \\
&=-\int^\pi_0 t~dt-\int^{2\pi}_{\pi}\underbrace{(t-2\pi)}_{\text{Principal value}}dt \\
&=-\int^\pi_0 t~dt-\int^{0}_{-\pi}u~du \qquad{(u=t-2\pi)}\\
&=\color{red}{0}
\end{align}
$$

