There is a result in complex analysis which says that a function which has no singularity in the finite part of the plane or at infinity is constant. Is this result applicable to the function $e^{-1/z^2}$?
A function with no singularity
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complex-analysis
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0No singularity ? https://www.wolframalpha.com/input/?i=plot+e%5E(-1%2Fz%5E2) – 2017-01-04
3 Answers
1
You can develop the function as the Laurent series
$$1-\frac1{z^2}+\frac1{2z^4}-\frac1{3!z^6}+\cdots$$
which shows an essential singularity at the origin.
3
You're wrong -- $f(z) = e^{-1/z^2}$ has an awesome singularity at $z = 0$. You are probably only thinking of the real-number version of this function, not the complex version.
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Your function is not defined at $z=0$ and cannot be continuously extended to $z=0$. To see this, consider the sequence $a_n=\frac in$, which converges to $0$ and notice that $$e^{-\frac1{a_n^2}}=e^{n^2}\overset{n\to\infty}\longrightarrow\infty.$$