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According to Wikipedia,

A zero of a meromorphic function $f$ is a complex number $z$ such that $f(z) = 0$. A pole of $f$ is a zero of $1/f$.

Is there a reason why a pole cannot be defined as the location where $\lvert f(z) \rvert = \infty$? I was imagining this to be the case based on this plot of the magnitude of the complex gamma function:

enter image description here

mhdadk
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3 Answers3

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Since $f$ is a function from $\Bbb C$ to $\mathbb C$, the range of $f$ cannot contain $\infty$. This is because $\infty\notin \mathbb C$, by definition. So, it does not make sense to say $|f(z)| =\infty$ (the modulus takes values in $[0, \infty)$ only).

Intuitively, what you say is true. In fact,

Suppose $f:\Bbb C\to\mathbb C$ has an isolated singularity at $a$. Then, $a$ is a pole of $f$ if and only if $$\lim_{z\to a} |f(z)| =\infty$$

stoic-santiago
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  • So can poles then be defined as the locations where $$\left\lvert \frac{1}{f(z)} \right\rvert = 0$$? – mhdadk Oct 19 '21 at 05:54
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    @mhdadk That is precisely what the definition you have stated (from Wikipedia) says. A pole of $f$ is a zero of $1/f$. We know that $$\left| \frac{1}{f(z)}\right| = 0 \Longleftrightarrow \frac{1}{f(z)} = 0$$ – stoic-santiago Oct 19 '21 at 05:55
  • The definition states that $1/f$ is $0$ and not $\lvert 1/f \rvert$. I consider these to be different, since $1/f$ can be complex while $\lvert 1/f \rvert$ is purely real. Is this reasoning correct? – mhdadk Oct 19 '21 at 05:56
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    @mhdadk, a complex number is zero if and only if its absolute value is zero. – YiFan Tey Oct 19 '21 at 05:57
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    @YiFan can't believe I missed that. Thanks. – mhdadk Oct 19 '21 at 05:59
  • @epsilon-emperor by the way, is there a reference for the statement you mentioned at the end of your answer? The one about "isolated singularity at $a$"? – mhdadk Oct 19 '21 at 06:00
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    Stein and Shakarchi's Complex Analysis would certainly have it somewhere in the chapter on Meromorphic Functions. @mhdadk – stoic-santiago Oct 19 '21 at 06:01
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In some sense the answer is no, but the reason is somewhat subtler. The issue is that a singularity of a general complex function $f$ is essentially a point where its absolute value is infinity, however, poles are only one particularly well-behaved sort of singularity, and not all singularities are poles.

For instance the function $$f(z)=e^{1/z}$$ is extremely poorly behaved at $z=0$, which you can sort of see by looking at a graph for real $z$. This is an example of an essential singularity, which is a singularity that is not a pole.

Note that this does not contradict the result mentioned in the other answer, because the limit $\lim_{z\to 0} e^{1/z}$ does not exist. Essentially, although $|f(0)|$ "should be" infinity, the limit does not actually exist because, if you pick just the right direction to approach zero (the imaginary axis), then the function actually does not blow up.

In fact, functions can have even worse singularities when they're clumped together, and these are certainly not poles. Examples of these are natural boundaries. You can see more information at the Wikipedia page.

YiFan Tey
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I'd say that your assertion is approximately correct, since one characterization of a pole of $f$ at $z_o$ (as opposed to essential singularity) is that $\lim_{z\to z_o} |f(z)|=+\infty$. In fact, although, there are complications in talking about "the value $\infty$", it can be made to make sense for poles: the function takes values on the Riemann sphere $\mathbb C\mathbb P^1$, which is obtained by adding (in suitable fashion) a "point at infinity" to the complex plane $\mathbb C$.

The behavior of $f$ with isolated singularity that is not a pole, but is an essential singularity, is not only captured by the negation of the condition for a pole, but, further, by the Casorati-Weierstrass theorem: in every neighborhood of an essential singularity $z_o$ of $f$, $f$ comes arbitrarily near to every value in $\mathbb C$.

paul garrett
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