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I have definition that holomorphic $f$ has singularity $z=a$ which is a pole of order $m$ iff its Laurent series at $a$ has zero coefficients for $n<m<0$ and $a_m \neq 0$. Does $\lim_{z\to 0} f(z) = \infty$ imply that $0$ is a pole?

3dok
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  • It may be an essential singularity. – Mark Fantini Jan 20 '14 at 05:57
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    @Fantini: I don't think so. The limit wouldn't exist if it were an essential singularity, right? Wouldn't $f$ assume all but finitely many values in every neighborhood of an essential singularity? – MPW Jan 20 '14 at 06:03
  • Yes, I recall that. But what is not existing? I'm thinking of the case $\exp(1/z)$. – Mark Fantini Jan 20 '14 at 06:08
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    @Fantini: Yes, that would have an essential singularity at $0$. But a nonconstant polynomial function of $1/z$ would not. In the first case the limit at 0 would not exist, while in the second case it would be $\infty$. There's a difference. For essential singularities, the function values "oscillate" wildly in every neighborhood of the singularity. For poles, they simply escape to infinity. A reasonable analogue on the real line might be $\sin 1/x$ versus just $1/x$. – MPW Jan 20 '14 at 06:16
  • Note that a limit cannot equal infinity, as infinity is not a number. The limit does not exist, but the function tends to infinity. – Glen O Jan 20 '14 at 06:51
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    @Glen O: It's conventional to write $\lim_{z\rightarrow z_0} f(z) = \infty$ if $f(z)$ can be made to lie in a given neighborhood of $\infty$ by taking $z$ sufficiently close to $z_0$. A neighborhood of $\infty$ is the exterior of a closed disk centered at $0$. Moreover, it is also conventional to think of maps as taking value in the Riemann sphere when convenient, so that $\infty$ is a point in that space. Admittedly, the domain/range were not explicitly mentioned here. – MPW Jan 20 '14 at 07:11
  • But $1/z$ is not a polynomial function. You are right though, functions oscillate wildly near essential singularities. – Mark Fantini Jan 20 '14 at 08:14

2 Answers2

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If it is isolated then $0$ is a pole see Cassorati-Weierstrass theorem

user68061
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  • I think, 3dok forgot to add the assumption that the function is holomorphic in a punctured neighborhood of zero, since otherwise, the notion of a pole is not even well-defined. In general, one can ask if under his/her condition $f$ admits a homomorphic extension to a punctured neighborhood of $0$. The answer to this would be negative. – Moishe Kohan Jan 20 '14 at 06:44
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it may be a pole bcoz we find poles of a function f(z) by equating denominator to zero

1.In case of rational functions if it contain some power of z say m in denominator & the powers of z present in numerator is either less than m or zero then $z=0$ is a pole

2.In case of trigonometric functions present in denominator we should find limit point of poles & they are 'non isolated essential singularity'.

3.There are some exceptional functions as i have one which is

$f(z)=exp(\frac{-1}{z^2})$

has no any singularity. as when $exp(\frac{1}{z^2})=0$

this equation does not hold for any z.

if any one has some more information & correction MOST WELCOME.

Community
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