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Let z=a+ib
So you can map the complex plane onto the surface of a sphere. Pretty neat. What if we cut the complex plane in half and only look at numbers with a negative real part?
Clearly, this subspace can be mapped to a half-sphere. It makes sense topologically too, as the subspace has one edge where a=0, and a half sphere only has one edge.
The surface of a half sphere, as I understand, is topologically equivalent to a circle. Does that imply that you could map the subspace of the complex plane onto a circle?
How would you go about finding such a mapping? Essentially, I want to flatten out this half sphere.
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OP here.
Just thinking, if you can map the half space where a<0 to a circle, surely you can map the half space where a<1 to a circle in a similar way. Or more generally the half space where a<n. Could you then let n tend to infinity and map the entire complex plane onto a circle?
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>>8538655
>How would you go about finding such a mapping?
w=(z+i)/(z-i)
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>>8538655
>The surface of a half sphere, as I understand, is topologically equivalent to a circle
I'm currently taking a course on complex analysis, so I'm not an expert, but, I think this statement is wrong.
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>>8539168
I have no formal background in topology but I don't see a problem with that.
Both the surface of a sphere and a filled circle are two-dimensional and have no holes so you should be able to homeomorph one into the other.
Also unlike OP I don't think it's that surprising that you can map the complex plane into a circle. We can already map the real line into the open interval (0,1), after all.
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>>8539168
Holy shit, the homeomorphism is literally right there in the OP pic, it's stereographic projection from the north pole.
The points that hit the disc and pass through the lower hemisphere should give you a hint,
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>>8538655
Chemistry student here, but I can't imagine laying out a Möbius onto a sphere in my head without intersection
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