Question about Riemann Surface Problem

[Crush1986]

Homework Statement

$$f(z)=z^\frac{3}{2}$$ find the branch points, branch cuts, and Riemann sheet structure.

Homework Equations

none

The Attempt at a Solution

So, I converted this to complex exponential form $$r^\frac{3}{2} e^\frac{i*3*\Theta}{2}$$ From here I mapped around a circle that was centered about the orgin. After cycling through 2 Pi I could see that z mapped into f(z) and wasn't at it's original point. So I concluded the branch points were at the origin and infinity (I think infinity is a branch point because z^-3/2 goes to 0 as z goes to infinity.

I think the cut can go from the origin to infinity in any direction.

The Riemann surfaces is giving me the most trouble. I keep going back and forth between this function being multivalued or not.

Thanks for any help!

 

[mighty2000]

Based on the given function, f(z) = z^(3/2), we can identify the branch points as z = 0 and z = ∞. This is because when z = 0, the function becomes undefined (division by zero) and when z = ∞, the function goes to infinity.

To find the branch cuts, we can look at the behavior of the function as we approach the branch points. As z approaches 0, the function becomes complex and has multiple values. Similarly, as z approaches ∞, the function also has multiple values. Therefore, we can draw a branch cut from 0 to ∞ in any direction, as you have suggested.

As for the Riemann sheet structure, it is important to note that the function f(z) = z^(3/2) is not single-valued. This means that there are multiple values of z that map to the same value of f(z). This is known as a multivalued function. In this case, the function has two branches, with branch points at 0 and ∞.

To visualize the Riemann sheet structure, we can use the mapping you have suggested, f(z) = r^(3/2) e^(i*3*θ/2). As you have observed, when we cycle through 2π, the point z maps back to itself, but the value of f(z) changes. This is because we are moving from one branch to the other.

To summarize, the Riemann sheet structure for f(z) = z^(3/2) has two branches, with branch points at 0 and ∞, and a branch cut from 0 to ∞ in any direction.