Find Continuity of f_α Along Negative X-Axis

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Homework Statement


Find a branch of w=z^{1/2} which is continuous along the negative x-axis.

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?
 
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Samuelb88 said:

Homework Statement


Find a branch of w=z^{1/2} which is continuous along the negative x-axis.

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?

There's a lot of choices for alpha. Why not pick alpha=0? Where is the discontinuity now?
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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