Complex Analysis: Fundamental Theorem of Calculus I was given the following question on a complex analysis exam but didn't answer it correctly:

Evaluate $$\int \sqrt{z-1} dz$$ about the the unit disk $|z|=1$ using only the Fundamental Theorem of Calculus. 

I confirmed numerically using mathematica that the answer is $-\frac{8}{3}i\ 2^\frac{1}{2}$.
Solution (with help of the lecturer):
The primitive $\frac{2}{3}(z-1)^\frac{3}{2}$ exists provided we have the branch cut $(-\infty, 1]$.
The path of the unit disk goes through the edge of the branch cut.
So let $z = e^{it}$ where $t=-\pi + \epsilon$ to $t=\pi - \epsilon$ and then take $\lim_{\ \epsilon \rightarrow 0}$
I obtain:
$$\frac{2}{3}(e^{i(\pi-\epsilon)}-1)^\frac{3}{2} - \frac{2}{3}(e^{-i(\pi-\epsilon)}-1)^\frac{3}{2}$$
But because the first term is above the negative real axis (the x axis) I get:
$$\frac{2}{3}(e^{i(\pi-\epsilon)}-1)^\frac{3}{2}\rightarrow \frac{2}{3}(2^\frac{3}{2})(-i)$$
Now the second term is below the negative real axis so 
$$\frac{2}{3}((e^{-i(\pi-\epsilon)}-1)^\frac{3}{2} \rightarrow \frac{2}{3}(2^\frac{3}{2})(+i)$$
So my main question is this:
Could someone please explain to me how these limits are evaluated (espically with respect to the branch cut)? 
Thanks in advance
Sorry about the type-setting as I am new to this website
 A: The choice of branch cut completely defines the problem. If we choose branch cut $[1,\infty)$ the $\int_{|z|=1} \sqrt{z-1}dz=0$ simply because the one point of non-analyticity (at 1) won't affect the integral. We must choose branch cut $(-\infty,1]$ (or similar) if we want a non-trivial answer. 
So to be very careful we should define our branch cut; specifically we define $$\sqrt{z-1}=\sqrt{|z-1|}e^{i\cdot Arg(z-1)/2},$$ where $Arg(z-1)$ is chosen to be in $(-\pi,\pi]$.
So, the evaluation of your statement by definition is, $$\frac{2}{3}\left(e^{(\pi-\epsilon)i}-1\right)^{3/2} = \frac{2}{3}\left(\sqrt{\left|e^{(\pi-\epsilon)i}-1\right|}e^{i\cdot Arg(e^{(\pi-\epsilon)i}-1)/2} \right)^3\to\frac{2}{3}\left(\sqrt{2}e^{i\cdot(\pi/2)}\right)^3=-\frac{4}{3}\sqrt{2}\cdot i$$ likewise, $$\frac{2}{3}\left(e^{(-\pi+\epsilon)i}-1\right)^{3/2} = \frac{2}{3}\left(\sqrt{\left|e^{(-\pi+\epsilon)i}-1\right|}e^{i\cdot Arg(e^{(-\pi+\epsilon)i}-1)/2} \right)^3\to\frac{2}{3}\left(\sqrt{2}e^{i\cdot(-\pi/2)}\right)^3=\frac{4}{3}\sqrt{2}\cdot i.$$ 
