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I wish to find what the Riemann surface diagram of $\dfrac1{\sqrt z}$ looks like.

The problem I'm having is that this function is not defined for $z=0$, and that this function doesn't map any $z\in\mathbb C$ to $0$.

Any hints ?

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    Not sure what you mean by "riemann surface diagram", but $1/\sqrt{z}$ has exactly the same "issues" as $\sqrt{z}$ (because the reciprocal part is not a problem), i.e., the function $1/\sqrt{z}$ can be defined on the same Riemann surface as for $\sqrt{z}$, excluding 0.2012-05-05
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    http://i.stack.imgur.com/PeIsB.png2012-05-05
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    To clarify, J. M.'s diagram is a picture of what _one sheet_ of the Riemann surface looks like. The whole Riemann surface is two copies of that, glued together along the edges in the obvious way. (What's not obvious is how to do it without self-intersection! I think it's impossible in $\mathbb{R}^3$...)2012-05-05
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    Hmm, that only was one sheet, wasn't it? [Let's fix that...](http://i.stack.imgur.com/htYb9.png)2012-05-05
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    Very nice, thanks!2012-05-05
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    Isn't this just a sphere? the genus should be 0.2012-05-05
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    @ChangweiZhou: It's not a projective variety. This is an open subset of the affine variety in $\mathbb{C}^2$ given by the equation $w^2 z = 1$.2012-05-06
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    @ZhenLin: I see. I confused this with the curve $z=w^{2}$.2012-05-07

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You may find this illustration and related discussion in this thread.
(this is an representation of the imaginary part, the real part is similar but the sheets are one over the other instead of interlaced as here)

1/sqrt(z+1)