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I understand (I think) how the Riemann surface of $\sqrt z$ is constructed, or even that of $\log z$. I can visualize them by transforming a plane according to the inverse of the function, i.e. $z^2$ for the former and $\exp z$ for the latter. See, for example https://www.youtube.com/watch?v=ZBvx4iEZOeA.

But then how do I construct the Riemann surface of $z^2$? In that case, the inverse function is multivalued which means that the transformation is not deterministic. So, should I split the points into two and end up representing them as two distinct points on the surface? As in -1 being transformed into both 1 and -1. Or, should I ignore one of the branches?

Sounds like a trivial question but I could not find any examples where the inverse of the function is multivalued.

Eser Aygün
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    I'm not experienced with Riemann surfaces, so I'll let someone else provide an actual answer, but it's my (very limited, and potentially incorrect) understanding that the point of using a Riemann surface $S$ for a *multi-valued* function was so that $f:\mathbb{C} \to S$ is single-valued and holomorphic. However $f(z)=z^2$ is not multi-valued, and is already holomorphic from $\mathbb{C}$ to $\mathbb{C}$, so why would you use a Riemann surface (other than $\mathbb{C}$)? – Joe May 22 '21 at 15:37
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    @Joe: That's a perfect answer. – Lee Mosher May 22 '21 at 15:55
  • My understanding was that every holomorphic function can be represented as a Riemann surface. $z^2$ is holomorphic therefore there must be a Riemann surface representing it. I just don't know how it looks.

    If that's too trivial, consider instead $\sqrt{(1 - z^3)}$. It is a multivalued holomorphic function whose inverse is also multivalued. So, how does one go about constructing its Riemann surface?

     – Eser Aygün May 22 '21 at 15:57
  • The Riemann surface for $f(z)=z^2$ is simply $\mathbb{C}$. If you want to know the Riemann surface for $f(z)=\sqrt{1-z^3}$, I recommend editing your question. – Joe May 22 '21 at 16:29
  • Isn't the deformation a part of the definition of the surface? Otherwise, the Riemann surface of $z^2$ would be equal to the Riemann surface of $z^3$, as they would both be simply $\mathbb C$. In any case, it seems like I need to do more reading on the subject before changing the question. – Eser Aygün May 22 '21 at 17:16

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