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I am reading some complex analysis and I am confused with the power function.

So I understand that $z^a$ has a branching point at $0$ if $a$ is not integer and that the number of branches can be infinitely many if $a$ is irrational. Then this means that $0$ is an essential singularity of the function and I actually computed few coefficients of the Laurent series of $z^\sqrt{2}$ as an example and it seems true.

So then, the big Picard theorem states that the function will take all possible values except possibly one infinitely many times. But on the other hand I find that $|z^a|=|z|^a$ for $a>0$, which means that not only there is a strict bound but also the limit to $0$ exists and it is $0$.

So I am confused as these cannot be both true. Where is my mistake?

rom
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    $0$ is not an essential singularity. The function does not exist in any punctured disk around zero. – Potato Nov 24 '13 at 00:48
  • What do you mean by saying that the function does not exist in any punctured disk around zero? Isn't zero a branching point? – rom Nov 24 '13 at 01:30
  • Still, even if you could define it. Why is $|z|^a$ bounded? You can make $|z|$ as large as you want. – user99680 Nov 24 '13 at 02:46
  • I mean it is bounded in a neighbourhood of zero which contradicts the big Picard theorem. – rom Nov 24 '13 at 03:00
  • What do you mean "even if you could define it"? Which thing cannot be defined? – rom Nov 24 '13 at 03:01
  • You do not have an isolated singularity. So your argument does not work. – Potato Nov 24 '13 at 03:53
  • A branch point is not an isolated singularity. Check the definitions again. – Potato Nov 24 '13 at 03:53
  • What? How is not zero an isolated singularity? Is there a sequence of sengularities converging to zero? – rom Nov 24 '13 at 03:56
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    So you mean zero is not an isolated singularity of $log(z)$? – rom Nov 24 '13 at 04:01
  • @rom Yes, because it is not isolated. You need to make a branch cut, which means you can't define it in a punctured disk around $0$, which means it isn't isolated. – Potato Nov 24 '13 at 17:52

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If you consider $z^\alpha$ (for $\alpha\notin\Bbb Z$) or $\log z$ as functions on $\Bbb C$ they do not have isolated singularities at $z=0$: there is no way to define them as holomorphic, or even continuous, functions on a punctured neighbourhood of the origin.

If you prefer, you can view those functions on (possibly infinitely sheeted) Riemann surfaces, and if you do so, the functions will indeed be holomorphic everywhere, but in that case, the origin will not lie on the Riemann surface in question, and it doesn't make sense to talk about the "singularity" at $z=0$.

In either way, $z=0$ is not an essential singularity, at least not with the common usage of the term. I have no idea how you "computed the Laurent series" for $z^\sqrt2$; that function does not admit a Laurent series around $0$.

mrf
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You are talking about $z^a$ which is $0$ at $z = 0$. When a function is zero at a point, that is not a singularity of any kind , even if it is also a branch point. All the branches of this function will be $0$ at $z = 0$; otherwise for the same z the different branches have different values.

A singularity is when the function goes to infinity as you approach the point, at least along some path. If the singularity is isolated the Laurent series will tell you what kind of singularity you have, by showing you which of the negative coefficients are $0$. If only a finite number of them are nonzero, you have a simple or multiple pole. If infinitely many of them are non-zero you have an essential singularity and can then apply the big Picard theorem.

Log z is a different issue. It does indeed have a singularity at $0$, and a branch point as well; but it is not an isolated singularity. You can read more about that here. Type of singularity of $\log z$ at $z=0$

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