Branch cut and contour integration along a special function Define for all complex $z$ except for a slit on the real interval $[0,1]$, the analytic function $f(z)=(z^2-z^3)^{-1/3}$, so that $f(z)$ is real valued on the upper side of the slit. 
a) How are the values of $f(z)$ on the lower side of the slit related to those on the upper side of the slit
b) Compute $\int_\gamma f$ taken once around any smooth curve $\gamma$ enclosing the slit. 
For a), should I pick the branch cut on the interval $[0,1]$. Namely, $0<\arg z<2\pi$ and $-\pi <\arg (z-1)<\pi$. Then we see that the difference between the argument is $\pi/3$ between the top and bottom of the slit.
For b), the gamma is taken around the interval $[0,1]$, so we should just use Cauchy integral formula correct? But this is different from the answer given. 
 A: First note that $f(z)=(z^2-z^3)^{-1/3}$ has branch points at $z=0$ and $z=1$.  Thus, we can choose to take as the branch cut the real-line segment $[0,1]$.  If we do so, then we must still define how the arguments of both $z^{-2/3}$ and $(1-z)^{-1/3}$ are defined.  
We can think of the branch cut (" the slit") as comprised of two branch cuts, one from $(0,0)$ to $(\infty,0)$ and the other from $(1,0)$ to $(\infty,0)$.  Alternatively, we could take the branch cuts to be from $(0,0)$ to $(-\infty,0)$ and the other from $(1,0)$ to $(-\infty,0)$.  

If we adopt the former convention for defining the arguments of both $z^{-2/3}$ and $(1-z)^{-1/3}$, then we have
$(i)$ On the "upper slit," the argument for $z$ is $0$ and the argument for $1-z$ is also $0$.  Thus, the argument of $f(z)$ is $0$.
$(ii)$ On the "lower slit," the argument for $z$ is $2\pi$ while the argument for $1-z$ remains $0$.  Thus, the argument for $f(z)$ is $-4\pi/3$.

If we adopt the latter convention for defining the arguments of both $z^{-2/3}$ and $(1-z)^{-1/3}$, then we have
$(i)$ On the "upper slit," the argument for $z$ is $0$ and the argument for $1-z$ is also $0$.  Thus, the argument of $f(z)$ is $0$.
$(ii)$ On the "lower slit," the argument for $z$ is $0$ while the argument for $1-z$ is $-2\pi$.  Thus, the argument for $f(z)$ is $2\pi/3$.  


IMPROTANT NOTE:
It is important to note that $e^{i2\pi/3}=e^{-i4\pi/3}$ and that the results for the two alternatives explored here are identical.


METHODOLOGY 1: Direct Integration of $f$ around $\gamma$
The integral $I$ around the contour $\gamma$ of the slit can be written
$$\begin{align}
I&=\oint_{\gamma}(z^2-z^3)^{-1/3}dz\\\\
&=\int_0^1 x^{-2/3}(1-x)^{-1/3}dx+\int_1^0e^{i2\pi/3}x^{-2/3}(1-x)^{-1/3}dx\\\\
&=\left(1-e^{i2\pi/3}\right)\int_0^1x^{-2/3}(1-x)^{-1/3}dx\\\\
&=\left(1-e^{i2\pi/3}\right)\text{B}\left(\frac13,\frac23\right)  \tag 1\\\\
&=\left(1-e^{i2\pi/3}\right)\frac{\Gamma\left(\frac13\right)\Gamma\left(\frac23\right)}{\Gamma\left(\frac13+\frac23\right)} \tag 2\\\\
&=\left(1-e^{i2\pi/3}\right)\frac{\Gamma\left(\frac13\right)\Gamma\left(1-\frac13\right)}{\Gamma\left(1\right)} \\\\
&=\left(1-e^{i2\pi/3}\right)\frac{\pi}{\sin(\pi/3)} \tag 3\\\\
&=\left(1-e^{i2\pi/3}\right)\frac{2\pi}{\sqrt{3}}\\\\
&=\pi(\sqrt{3}+i)
\end{align}$$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{I=\oint_{\gamma}(z^2-z^3)^{-1/3}dz=\pi(\sqrt{3}+i)}$$

Special Notes:
In $(1)$, $\text{B}(x,y)$ is the Beta Function.
In $(2)$, $\Gamma(x)$ is the Gamma Function, which is related to the Beta Function by $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$
In $(3)$, we used Euler's Reflection Principle, $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin (\pi x)}$.

METHODOLOGY 2: Using the Residue at Infinity to Integrate of $f$ around $\gamma$
Another way to evaluate the integral $I$ is to evaluate the Residue at Infinity.  The residue at infinity is given by
$$\begin{align}
\text{Res}\left(f,\infty \right) &= \text{Res}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right),z=0\right)\\\\
&=\lim_{z\to 0}\,z\,\left(-\frac{1}{z^2}\left(\frac{z-1}{z^3}\right)^{-1/3}\right)\\\\
&=-e^{-i\pi/3}
\end{align}$$
Thus, 
$$\bbox[5px,border:2px solid #C0A000]{I=-2\pi i (-e^{-\pi/3})=\pi(\sqrt{3}+i)}$$ 
as expected!
