Let $$f(z)=\sin\left(\frac{\pi}{z^2+1}\right)$$
How can I show that $z=i$ is an essential singularity?
I have tried many things like expanding, or derivate. But I can't get something. Some ideas or hint will be useful.
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Going from the wikipedia article, you can show that the limit as z approaches i of $f(z)$ and $\frac 1 {f(z)}$ both do not exist. – Alan Oct 07 '14 at 17:25
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but non-existence is not enougth to say that is an essential singularity. – Porufes Oct 07 '14 at 17:40
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Have you tried expanding the function into a Laurent series about $z=i$? – user_of_math Oct 07 '14 at 17:44
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I have but it's to hard – Porufes Oct 07 '14 at 17:57
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If $i$ is not an essential singularity, then there is an integer $k$ such that $$\lim_{z\to i}(z-i)^k f(z) = 0\tag1$$ hence $$\lim_{z\to i}(z^2+1)^k f(z) = 0\tag2$$
After the change of variable $\zeta=\pi/(z^2+1)$ equation (2) yields $$\lim_{\zeta\to \infty} \zeta^{-k} \sin \zeta=0 \tag{3}$$ But Entire function bounded by a polynomial is a polynomial.
