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I have a function,
$$ \frac{ e^{\sqrt{z}} - e^{-\sqrt z}}{\sin \sqrt z} $$

Does this function have removable singularities? to me it seems like there is a branch point at $z=0$. Thanks for any inputs.

user38249
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2 Answers2

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A related technique. Here is a start. $$ \frac{ e^{\sqrt{z}} - e^{-\sqrt z}}{\sin \sqrt z}= \frac{(1+\sqrt{z}-\dots)-(1-\sqrt{z}+\dots)}{\sqrt{z}-\frac{\sqrt{z}}{3!}+\dots}. $$

Can you see what's going on?

Community
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Mhenni Benghorbal
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  • So with that, I can divide numerator and denominator by $ \sqrt z $ that tells me the fact that this is a removable singularity !! ** So it doesn't have branch cuts ? ** – user38249 Jan 27 '14 at 16:05
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Let's start with the function $g(w) = (e^w-e^{-w})/\sin w$, which has a removable singularity at $w=0$. The question to address: is $g$ an even function? If it is, then yes, we can plug $\sqrt{z}$ into it and not worry about branches at all: the composition will be holomorphic near zero. (All powers in the Taylor series of an even function at $0$ are even.)

So: is $g$ even? I guess it is!