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Suppose $f$ is analytic in $\{0 < |z| < R\}$ and that the range of $f$ does not include the negative real axis. Prove that $f$ has either a removable singularity or a pole at $0$. (Don’t use the Picard theorem).

Hint: In this case we can define $\sqrt{f(z)}$ (Why?). Look at the range of $\sqrt{f(z)}$ and notice it omits a disc. Use the Casorati–Weierstrass theorem.

I feel like I'm being given most of the problem via this hint, but I'm still not seeing what I should be doing here. Why does omitting the negative real axis make a difference to whether I can define $\sqrt{f(z)}$? And why would the range of this function omit a disk?

I know that once I can see a disk is omitted from the range, this means that $\sqrt{f(z)}$ cannot have an essential singularity at $0$ because this would contradict the Casorati-Weierstrass theorem, I just don't see why this is true to begin with. I don't want somebody to just do the problem for me, I just want help seeing whatever it is that I'm missing here.

Alex Mathers
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1 Answers1

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If $f$ is non-constant, then it is an open map, hence if the range does not contain the negative real axis then it does not contain $0$ either. Therefore the range of $f$ is contained in $\mathbb{C}\setminus(-\infty,0]$, and hence $\sqrt{f(z)}$ can be defined via the principal branch of the logarithm.

Since this branch of $\sqrt{z}$ maps the slit plane $\mathbb{C}\setminus(-\infty,0]$ into the right half plane, it follows that the image of $\sqrt{f(z)}$ is contained in the right half plane, hence it certainly omits a disk.

carmichael561
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  • carmichael if i want to show that 0 is is removable singulairty when R=1 can i use the same function $\sqrt{f(z)}$ to show that the composition is bounded? – convxy Jun 01 '20 at 19:22