I am trying to calculate $\oint \sqrt{z} dz$ around the unit circle with $z = e^{2\pi it}$ from $t = 0$ to $t = 1$.
I realize that because $\sqrt{z}$ isn't analytic on the whole plane, there are some technical difficulties with doing this. But I am imagining taking a branch cut along the line from $0$ through $1$ out to $\infty$ and the endpoints of the integral can be taken as going from $e^{i\varepsilon}$ to $e^{(2\pi - \varepsilon)i}$ with $\varepsilon$ going to $0$. In the limit, this approaches the whole circle and because there isn't any asymptotic behavior at $z=1$, the limit should converge without any difficulties.
However, I'm unsure of my reasoning as I go about calculating this.
I start by saying $\oint \sqrt{z}dz = \int_0^1 \sqrt{e^{2\pi it}} d(e^{2\pi it}) = \int_0^1 e^{\pi it} e^{2\pi i t}\cdot 2 \pi i dt = 2\pi i \int_0^1 e^{3\pi it}dt$. I then take an antiderivative of the inside: $\frac{d}{dt} \frac{1}{3\pi i} e^{3\pi i t} =e^{3\pi it}$, and so I can evaluate the integral at the endpoints: $2\pi i \int_0^1 e^{3\pi i t}dt = 2\pi i \big[\frac{1}{3\pi i} e^{3\pi i t}\big]_0^1 = \frac{2}{3}(e^{3\pi i} - 1) = -\frac{4}{3}$.
I can't spot any immediate errors in my reasoning, but at the same time, it's loose enough that I'm not sure it's correct.
Is my answer correct? And is there any better way of approaching this problem?
Note that $\oint_\gamma\sqrt{z}\>dz$ with $\gamma$ the unit circle is undefined unless you specify the branch cut; hence this cut is not just a computational device! You have chosen the branch cut along the positive real axis, which is o.k. But we are not done yet: Your $\sqrt{z}$ can now be chosen as an analytic function in the the rest of the plane, but there are two choices possible, exemplified by $\sqrt{-1}=i$ and $\sqrt{-1}=-i$. Without much ado you have chosen the first of these and then went on with your calculation. Since the second choice differs from the first only by a factor $-1$ you would have obtained ${4\over3}$ choosing the other branch.