complex logarithm negative real axis The function $\log(z)=\log|z|+i \arg(z)$, is continuous in the domain $\mathbb{C}-\mathbb{R^-}$.
But  can we say that it's defined on $\mathbb{C^*}$.
Like for example : $\log(-1)=\log(\exp(i\pi))=i\pi.$ is it true ?
 A: Well, you can either choose your branch cut somewhere else (for example, along the negative imaginary axis) or you can form a Riemann surface (then the function $f(z)=\log z$ becomes multiple valued).  In theses cases, the function is defined along the negative real axis.  
If your branch cut is along the negative real axis, then your function is not defined there.
A: Before I answer your question, I'll make the note that you are writing $\log(z)$ on the left side of your equations, but are returning the principal value of the $\log(z)$ function on the right side. The principal value has the branch cut, $\log(z)$ does not. This was denoted in my textbook as $\text{Log}(z)$, or $p.v. \log(z)$. 
Even the principal value is defined everywhere on $(-\pi, \pi]$. So yes, you are correct $\text{Log}(-1)=i\pi$. The problem arises when you try to integrate the value. Because, 
$$\lim_{\epsilon\to0}\left(\text{Log}(-1-i\epsilon)\right)=-i\pi$$
which is off by $2i\pi$ from the result of $\text{Log}(-1)$. This is problematic when we try to integrate, so we cannot integrate through this branch cut. However, the function is defined on the negative real axis, it just becomes discontinuous. 
