Contour Integral involving a branch of $1 / \sqrt{z}$

Question

I was asked the following question involving computing contour integral for a certain branch of the complex function $1 / \sqrt{z}$.

This is mainly about testing our understanding of different branches of complex Log(z) function (and hence of the appropriately defined complex square root function), so I'm looking for feedback as to whether i'm interpreting the data correctly.

We are given:

1. $\Gamma$ is the directed contour corresponding to the half unit circle $\{ z: Imz \ge 0, |z| = 1 \}$ in the positive (anti-clockwise) direction.
2. Mark by $\sqrt{z}$ the branch of $(z)^{1/2}$ defined on the region $\{ z: Argz \neq - \frac{\pi}{2} \}$ so that $\sqrt{1} = 1$.

We need to compute: $$\int_{\Gamma}\frac{dz}{\sqrt{z}}$$

My solution:

In our course, we define the complex power function as $z^a = exp(\alpha L(z))$, where $L(z)$ corresponds to the branch of the complex logarithm defined for our domain. In this case per the data, we are interested in the branch $L(z)$ which is defined everywhere on the complex plane except for the non-positive imaginary axis ($\{ z: Argz \neq - \frac{\pi}{2} \}$).

Now, even within this branch a square root has two solutions, differing by $i \pi$ from each other. So the data point $\sqrt{1} = 1$ is what helps us unambigously 'pin down' the particular function we are interested in. (Is this paragraph right?).

So for a number on the unit circle, $e^{it}$, the function takes the form $e^{i\frac{1}{2}t}$.

Parameterizing appropriately we can therefore compute the integral using $\gamma(t) = e^{it}$ where $t \in [0, \pi]$ and $\gamma'(t) = ie^{it}$ to get:

$$ \int_{\Gamma}\frac{dz}{\sqrt{z}} = \int_{\gamma}\frac{dz}{\sqrt{z}} = \int_{0}^{\pi}\frac{1}{e^{0.5it}} ie^{it}dt = i\int_{0}^{\pi}e^{0.5it}dt = ... = 2(i - 1)$$

Is this right? If we were given for example that for our current choice of this function it holds that $\sqrt{1} = -1$ how would that change the picture? I suppose in that case we define it to be something like $\sqrt{e^{it}} = e^{i(\frac{1}{2}t + \pi)}$, but then it doesnt really affect the integral value ultimately?

Answer 1

Yes it is correct. If you take the other branch you get the same but with a minus sign in front because the integrand in that case is just the negative of the integrand you had in the branch before.