# Branch of z^{1/2}

1. Oct 9, 2011

### Samuelb88

1. The problem statement, all variables and given/known data
Find a branch of $w=z^{1/2}$ which is continuous along the negative x-axis.

3. The attempt at a solution

The book proves that the principle square root function $|z|^{1/2} \big( \cos(\theta/2) + i \sin(\theta/2) \big)$, where $-\pi < \theta \leq \pi$ is discontinuous along the negative x-axis.

I've defined a new branch of the square root function $f_\alpha (z) = |z|^{1/2} \big( \cos(\theta/2) + i\sin(\theta/2) \big)$, where $\alpha < \theta \leq \alpha + 2\pi$. I know the principle square root function is discontinuous along the negative x-axis because the limit as the principle square root function is path dependent as $(r,\theta)$ approaches an arbitrary negative number $r_0 e^{i \theta_0}$. What's the best way to proceed from here? Should I choose an $\alpha$ such that $f_\alpha$ can only approach the negative x-axis one way?

2. Oct 9, 2011

### Dick

There's a lot of choices for alpha. Why not pick alpha=0? Where is the discontinuity now?

3. Oct 9, 2011

### Samuelb88

Along the positive x-axis. I understand how I should approach the problem now. Nonetheless, thank you for your help! My book hid the fact that branches are discontinuous along their branch cuts at the end of an example.