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I'm new to the topic of Riemann surfaces and complex analysis, and I tried to compute the Riemann surface of $z^3$ on Wolfram Alpha but couldn't because its not a multivalued function. How can I compute the riemann surface of $z^3$ or I got something really wrong?

user773458
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Krcx
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When thinking of the Riemann surface of an expression with one complex variable, you should really think of converting that into an equation with two complex variables, like this: $$w = z^3 $$ The resulting Riemann surface is then the solution set of this equation: $$S = \{(w,z) \in \mathbb C^2 \mid w=z^3\} $$ However, as you have observed, $w=z^3$ is single valued as a function of the variable $z$. So that equation defines a function with independent variable $z$ and dependent variable $w$. Also $S$ is simply the graph of this function (with domain--codomain variables switched from their usual "first--last" positions). So as is usual with functions, we obtain an isomorphism between the domain of the function and the graph of the function, namely $\mathbb C \mapsto S$ given by $z \mapsto (w,z^3)$. In this context "isomorphism" means "biholomorphism" and so the Riemann surface is (biholomorphic to) $\mathbb C$.

Lee Mosher
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  • But still there must be a sheet that contains the results of the equation right, because I've seen someone plotting a Riemann surface for w=z^2+1. – Krcx Feb 04 '24 at 13:52
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    OP wrote $z^3$, and you wrote $z^2$, but what you wrote applies to $z^3$ as well – J. W. Tanner Feb 04 '24 at 13:54
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    Oops, thanks for that – Lee Mosher Feb 04 '24 at 13:57
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    Now what happens if we consider $S$ the other way around ... $w=z^3$ thought of as $z = w^{1/3}$. In this case, the same set $S$ and the same Riemann surface is thought of as a $3$-fold covering of the $w$-plane. When you say "the Riemann surface of $z^3$", perhaps that is what is intended. – GEdgar Feb 04 '24 at 13:57
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    @Krcx: Regarding your comment, $w=z^2+1$ also defines $w$ as a function of $z$, and the domain of that function, namely the $z$-plane, is also (identified with) the Riemann surface. Your words regarding a sheet that contains the results of the equation are rather vague and I'm not sure how to address them, but whatever those words mean, the point is simply that in the case of an equation $w=f(z)$, the $z$-plane is the thing you want. – Lee Mosher Feb 04 '24 at 14:01
  • Much thanks to all of you, I am new to the topic and trying to learn it on my own (11th grade) I haven't though of the possibility of switching between w-plane and z-plane therefore I was confused. – Krcx Feb 04 '24 at 14:04