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I am stuck on a problem which is given as an exercise in my book. I really would like to solve it primarily by myself, but unfortunately I don't "see" the solution intuitively and therefore I have no good starting point. The problem is the following: Show that if $f$ has an isolated pole singularity at $c$, every neighborhood of $c$ that lies in the domain of f contains all complex numbers $z$ with $|z|>r$ , for some $r$ dependent on the choice of neighborhood. A small hint would be appreciated. It will allow me to get some sleep.

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2 Answers 2

Use the definition of pole and the fact that holomorphic functions map open sets to open sets. Full solution is written below.


By definition, if $(1/f)(c)$ is defined to be zero, $1/f$ is holomorphic at $c$ and has a zero there. Since holomorphic functions map open set to open sets, image of a neighborhood of $c$, say $V$, under $1/f$ contains an open disc centered around $0$, say with radius $R$. This means that image of $V$ under $f$ contains all points whose distance from origin is farther than $1/R$.

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You should simpy consider $1/f$. This function is holomorphic around $c$ and also at $c$ with $f(c)=0$.

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