I have a slight confusion regarding poles/zeros and I'd like to clear things up.

$f$ has a pole is $1/f$ has a zero and vice versa:

Do polynomials have poles?

My answer is no, they can only have zeros.

But what if I take the polynomial $z^n+1=\frac{z^{n+1}+z}{z}$. This looks like it has a pole at $z=0$

  • $\begingroup$ It doesn't, you can see it in the Laurent series, which is the first form you wrote. $\endgroup$ – HEXQ Mar 27 '16 at 19:38
  • $\begingroup$ It has really not a pole. If you take the $\lim_{z\to0}$ of your last expression, you will realize that $z=0$ is not really a pole. A polynomial can never has poles $\endgroup$ – seoanes Mar 27 '16 at 19:39
  • $\begingroup$ @hrt So we always need to reduce the function, cancel everything $\endgroup$ – GRS Mar 27 '16 at 19:39
  • $\begingroup$ @GRS If you can yes, you won't make a pole disappear by simplifying correctly and you won't create a pole by multiplying the numerator and denominator by $z$. Polynomials are equal to their Laurent series, so you can see immediately that they won't have any pole because the singular part is always zero. $\endgroup$ – HEXQ Mar 27 '16 at 19:45
  • $\begingroup$ @ GRS : a general fact when we consider (locally) holomorphic functions, is that the behaviour of $f(z)$ at some point $a$ is fully determined by the values of $f$ in the neighborhood of $z=a$. hence, when you consider $z^n+1$ and $\frac{z^{n+1}+z}{z}$ at $z= 0$, you don't care of if they are defined or not at $z=0$ since you only look at their values in the neighborhood of $z=0$ $\endgroup$ – reuns Mar 27 '16 at 19:46

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