analytic continuation for $(e^z)^{\frac{1}{3}}$ We have the function $(e^z)^{\frac{1}{3}}$
Choose a branch that is analytic in the circle $|z-2| <1$. Then analytically continue this branch over the curve $\gamma$, where $\gamma$ a curve going from $z=2$ around origin counter-clockwise in a loop back to $z=2$. Do the new functional values agree with the old ones?
So i did this already for the function $3z^{\frac{2}{3}}$, all i did was to rewrite it in exponential form and then going around the loop once will add some radians to the argument making the functional values not to agree.
How do i do it for this exponential function? Im not even sure how to define a branch for this, any hints? 
 A: The analytic continuation theorem states that two holomorphic functions on a domain which agree on an open subset of that domain are equal. (The statement can be made stronger but we don't need more.) 
Therefore any function $f$ which is holomorphic on your disk and satisfies $f(z)^3 = e^z$ is actually equal to $e^{z/3}$ on all of $\mathbb C$ (see Edit : or a third root of unity multiple of it) since the latter is holomorphic on all of $\mathbb C$.
EDIT : As was obviously pointed out, I mistook $e^{z/3}$ for $(e^z)^{1/3}$.  The point is extracting the third root last involves a choice of branch for $g(w) = w^{1/3}$. Choose a branch for that function and then compose it with $e^z$ ; you should obtain either of $e^{(z+2*k*\pi)/3}$ for $k=0,1,2$, which are three distinct branches that are analytic over all of $\mathbb C$ and agree nowhere ; note that $e^z \neq 0$ for all $z \in \mathbb C$, which is the branch point of $g$ ; this explains why the branches are holomorphic everywhere.
Hope that helps,
