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this is my first post so sorry if my question is too vague. I didn't see a related question posted, hence why I'm asking.

I can't find any resources on it, but there's supposed to be a bijective mapping from the complex plane to the reimann sphere, correct? The only mapping I've seen is by creating a line from a point in the complex plane to the top of the sphere and the intersection point with the sphere is the function value. How can this be expressed, and how is it injective?

On a intuitive note, how can the complex plane be isomorphic to a sphere? I'd think you could 'unfold' the sphere which would create a finite plane hence not being injective.

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  • $\begingroup$ Hello and welcome to MSE. I believe the stereographic projection is what you are looking for. I believe that the wikipedia article provides a nice explanation, and there's a lot of youtube videos explaining this as well. $\endgroup$ – Alexandros Feb 15 at 22:10
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Given a complex number $z=x+iy$ we think of it as a vector in $\mathbb{R^3}$ as $(x,y,0)$. There is a unique line which passes through this point and through the north pole $(0,0,1)$. It is given by the equation $f(t)=(1-t)(0,0,1)+t(x,y,0)=(tx,ty,1-t)$. We want to find where does this line intersect the sphere. So let's see when $||f(t)||^2$ equals to $1$.

$||f(t)||^2=t^2x^2+t^2y^2+(1-t)^2=t^2x^2+t^2y^2+1-2t+t^2=t^2(x^2+y^2+z^2)+1-2t$

So let's compare this to $1$. We get $t^2(x^2+y^2+1)+1-2t=1$ and from here $t^2(x^2+y^2+1)=2t$. One possible solution is obviously $t=0$ and then the point of intersection with the sphere we will get is the north pole. But we are interested in the other point of intersection. So let's suppose $t\ne 0$ and divide by $t$. Then we get $t=\frac{2}{x^2+y^2+1}$, so the point of intersection is:

$(tx,ty,1-t)=(\frac{2x}{x^2+y^2+1},\frac{2y}{x^2+y^2+1},\frac{x^2+y^2-1}{x^2+y^2+1})=(\frac{2Re(z)}{|z|^2+1},\frac{2Im(z)}{|z|^2+1},\frac{|z|^2-1}{|z|^2+1})$

Alright, so we finally got the required formula. And it has an inverse map from $S^2\setminus\{(0,0,1)\}$ to $\mathbb{C}$ given by $(\xi,\eta,\lambda)\to \frac{\xi+i\eta}{1-\lambda}$ as you can check. So it is a bijection between the complex plane and the sphere without the north pole. Note that it is very important we exclude the north pole.

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