MHB Plotting Complex Functions: Does it Look Like a Riemann Surface?

Dustinsfl
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I plotted the real and imaginary parts of a complex function \(z^{1/3}\). The two plots are similar to the Riemann surface is that correct?

twlfDg3.png


qTfSwPj.png
 
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Recall that the Riemann surface of $w^3 = z$ is defined to be

$$\{(w_0, w_1, z_0, z_1) \in \Bbb R^4 \, :\, (w_0 + iw_1)^3 = z_0 + iz_1\}$$

While taking the real part you are really plotting

$$\{(x, y, z) \in \Bbb R^3 \, : \, z = \Re[(x+iy)^{1/3}]\}$$

So that's bound to be a section of the original Riemann surface, a bit deformed around the branches. Similar holds for the imaginary part.

So yes, those indeed look similar to the original Riemann surface.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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