If $C$ is a positively oriented circle$ |z|=1$then by cauchy integral formula $\int_C \frac{dz}{z} =2\pi i$
Decompose$ C$ in the form $C=C_1+C_2$, Where $C_1$ is the upper half of C from $1$ to $ -1$ and $C_2$ is lower half of$ C$ from$ - 1$ to $1$, then consider the following argument and find the fallacy which leads one to conclude that $0=2\pi i$.
$\int_C \frac{dz}{z}=\int_{C_1} \frac{dz}{z}+ \int_{C_2} \frac{dz}{z}= \int_{1}^{-1} \frac{dz}{z}+ \int_{-1}^{1} \frac{dz}{z}$= $Log (z) |_{1} ^{-1}$+$Log (z) |_{-1} ^{1}$=$Log(-1)-Log(1)+Log(1)-Log(-1)=0$
I have tried so hard to find the fallacy but couldn't ... So please help me. Thanks in advance.
When you decomposed it to $C_{1}$ and $C_{2}$ your integral is still pretty simple to solve:
All complex numbers on your countour $C_{1}$ can be written as $z=re^{i\theta}$, where $r=1$ and $0\leq\theta\leq\pi\implies z = e^{i\theta}$ on this contour. Futhermore, $dz= ie^{i\theta}d\theta$
Now, $\int_{C_1} \frac{dz}{z}= \int_{0}^{\pi} \frac{ie^{i\theta}d\theta}{e^{i\theta}}=i\int_{0}^{\pi}d\theta =i(\pi-0)=\pi i$
Similary for $C_{2}$, we have that $z=e^{i\theta}$ on this contour, where $\pi\leq\theta\leq2\pi$
Our integral on this contour becomes: $\int_{C_2} \frac{dz}{z}= \int_{\pi}^{2\pi} \frac{ie^{i\theta}d\theta}{e^{i\theta}}=i\int_{\pi}^{2\pi}d\theta =i(2\pi-\pi)=\pi i$
Add them both up to get $2\pi i$, which is consistent with your first method!
As for your method, the $Log$ function is usually defined to have a branch cut on the negative real axis so it can be single valued. I'm not sure how you've defined it.