principal roots

Question

When we define the principal square root of a number like so: $$\sqrt{x}=\sqrt{r}e^{ \frac{i\theta}{2}}$$ am i right that the definition basically depends on the range we choose for $\theta$? as a range from $-\pi$ to $\pi$ will give a different function than one from $0$ to $2\pi$ say. I believe convention is $-\pi$ to $\pi$ with $\pi$ being inclusive. i would like to know if this represents a branch cut even though the axis is lumped with one side of the inequality

Answer 1

Yes, when you're defining a single-valued version of the two-valued square-root function in this way, the range you choose for $\theta$ corresponds to which version of the function you are choosing. What's usually called principal branch is when you choose $\pi<\theta \le \pi.$ This corresponds to choosing the branch cut along the negative real axis, since there is a discontinuity there: points just below the negative real axis get sent toward the negative imaginary axis and those just above get sent toward the positive imaginary axis.

The choice of $\theta \le \pi$ as the inclusive side means that the negative real axis is sent to the positive imaginary axis, so that, for instance $\sqrt{-1} = +i.$ This choice is pretty immaterial: we could have easily sent $-1$ to $-i$ instead by choosing $-\pi \le \theta < \pi$ as our range for $\theta$. The first is just the conventional choice for the 'principal' branch. And either way, we say there's a branch cut on the negative real axis.

If instead we choose $0\le \theta < 2\pi$ then instead there is a branch cut on the positive real axis. Points on and just above the real axis get sent toward the positive imaginary axis and points just below get sent toward the negative imaginary axis.

Answer 2

If $-\pi\le\theta\le\pi$ then via an Argand Diagram, your principal value of the square root covers all $z\in\Bbb C$ such that $\Re(z)\ge0$

On the contrary, if $0\le\theta\le 2\pi$, the definition extends to all $z\in\Bbb C$