Making a cube root function analytic on $\mathbb{C}\backslash [1,3]$ I am still not convinced by the post that the function$$\sqrt[3]{(z-1)(z-2)(z-3)}$$ can be defined so it is analytic on $\mathbb{C}\backslash [1,3]$. We define for each $z\in \mathbb{C}\backslash (-\infty,3]$ the function $$f(z)=\int_4^z \frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}\,dz +\ln 6$$ and it claims that function $$\exp\left(\frac{1}{3}f(z)\right)$$ is continuous on $(-\infty,1)$. This is the part I don't understand. How do we prove that it is continuous on $(-\infty,1)$. I computed the integral from 4 to $-4$ along the upper semicircle of $|z|=4$, and the integral from 4 to $-4$ along the lower semicircle of $|z|=4$ and they are not equal. So I don't see how the function can be continuous at $-4$, for example.
 A: Given that
$$\frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}=\frac1{z-1}+\frac1{z-2}+\frac1{z-3}$$ you see that all the three residues of the integrand are equal to one, so their sum is $3$. The "upper" and "lower" integrals thus differ by $6\pi i$.
After multiplication by $1/3$ they differ by $2\pi i$, and after $\exp$ they agree. IOW $f(z)$ won't be continuous, but $\exp(f(z)/3)$ still is.
A: If you  believe in the Monodromy theorem and are comfortable with the Riemann sphere $\overline{\mathbb C}$, the following short argument suffices: 
$$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = z\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$$
Here $\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$ admits analytic continuation in $\overline{\mathbb C}\setminus [1,3]$, which is a simply-connected domain. Therefore, it has a single-valued branch there. 

If you are not comfortable with the Riemann sphere, but still  believe in the Monodromy theorem, then move the segment to infinity, for example by letting $w=1/(z-1)$. This transforms $\overline{\mathbb C}\setminus [1,3]$ into $ {\mathbb C}\setminus [1/2,\infty)$. We get 
$$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = \Big(  w^{-1}(w^{-1}-1) (w^{-1}-2) \Big)^{1/3}
=w^{-1} ( (1-w)(1-2w) )^{1/3}
$$
Here $( (1-w)(1-2w) )^{1/3}$  admits analytic continuation  in $ {\mathbb C}\setminus [1/2,\infty)$, which is a simply-connected domain. Therefore, it has a single-valued branch there. 
