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Suppose $f(z) → \infty$ as $z → z_0$, so there's an isolated singularity at $z = z_0$. Ultimately I intend to show that there is a pole at $z_0$? I realize this has been asked before, but my question is different because I am asking specifically about the approach outlined below and whether or not the converse of the following two thereoms are true.

I know that I can do this by showing that $z=z_0$ is neither a removable singularity nor an essential singularity. However, I'm not positive why it's not either of those.

I know that if $f$ is bounded in a deleted neighborhood of an isolated singularity, then the singularity is removable. But I don't know if the converse is true. Is it? Also I know that if $f$ has an isolated singularity at $z_0$ and if lim$_{z→z_0}(z-z_0)f(z) = 0$, then the singularity is removable. But again I don't know if the converse is true. Is it? To show it's not an essential singularity, I can just use the Casorati-Weierstrass Theorem and show that if $D$ is a deleted neighborhood of $z_0$, then the range $R = {{f(z):z\in D}}$ is not dense in the complex plane. But how would I show that this is true ?

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  • $\begingroup$ What is your definition of "pole"? ("$f(z)\to\infty$ as $z\to z_0$" is often taken as the definition...) $\endgroup$ Commented Dec 5, 2015 at 22:10
  • $\begingroup$ My definition is that for $z$ not equal to $z_0$, $f$ can be written in the form $f(z) = A(z)/B(z)$ where $A$ and $B$ are analytic at $z_0$, $A(z_0)$ is not equal to $0$, and $B(z_0) = 0.$ $\endgroup$
    – Jimm
    Commented Dec 5, 2015 at 22:17

2 Answers 2

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If $f$ has a removable singularity at $z_0$, then we can extend $f$ to $z_0$ to take a value $f(z_0)=a$ so that it is an analytic function at $z_0$. In particular, it is continuous at $z_0$, so by definition of continuity, for any $\epsilon>0$ there is a neighborhood of $z_0$ on which $|f(z)-a|<\epsilon$. In particular, $|f(z)|< |a|+\epsilon$ is bounded on such a neighborhood. Similarly, by continuity $\lim_{z\to z_0}(z-z_0)f(z)=(z_0-z_0)a=0$.

Also, if $f(z)\to\infty$ as $z\to z_0$, then for any $N$, there is a deleted neighborhood of $z_0$ on which $|f(z)|>N$, and then the image of that neighborhood cannot be dense as long as $N>0$ (because the image is contained in the closed set $\{w:|w|\geq N\}$).

But a simpler approach to this problem (which does not require the full classification of isolated singularities) is just to consider the function $g(z)=1/f(z)$. Since $f(z)\to \infty$ as $z\to z_0$, $g(z)\to 0$ as $z\to z_0$, so $g$ has a removable singularity at $z_0$ with value $0$. Filling in that singularity, we can write $f(z)=1/g(z)$, where $1$ and $g(z)$ are both analytic functions at $z_0$, with $g(z_0)=0$ and $1(z_0)\neq 0$. Thus $f$ has a pole at $z_0$.

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  • $\begingroup$ Thank you very much, this was very clear and explained it exactly $\endgroup$
    – Jimm
    Commented Dec 5, 2015 at 22:40
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What if $f(z) =\exp(\frac1{|z-z_0|}) $?

There weren't any restrictions on $f$ that I could see.

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