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I want to find the order of the pole of $\sqrt{1+\frac{1}{z}}$ at 0. But how can I expand this function at 0? Can I take the Taylor expansion of $\sqrt{1+z}$ and plug in $\frac{1}{z}$ in the place of z? And how can one find the Laurent series expansion of a function in general? I mean since the function diverges toward infinity at poles, how can the Laurent series be "defined"?

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  • $\begingroup$ The function $\sqrt{1+1/z}$ has branch points at $z=-1$ and $z=0$. So, you can develop the Laurent series outside the unit disk. $\endgroup$
    – Mark Viola
    Jan 17 at 16:28
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You're quite right that to find the Laurent series expansion for $f(z)=\sqrt{1+\frac1z}$ about $0,$ we need only start with the Taylor series expansion of $g(z)=\sqrt{1+z}$ about $0,$ then note that $f(z)=g(1/z).$ In fact, this will show you that it isn't a pole, but rather an essential singularity.

As for how the Laurent series can be thought of as being defined, recall that a power series is defined wherever it converges. In this case, it converges at all values of $z$ such that $|z|>1.$

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    $\begingroup$ The Taylor series of $g$ around $z = 0$ converges for $|z| < 1$, so the corresponding Laurent series of $f$ converges for $|z| > 1$, not for all non-zero $z$. $f$ does not have a Laurent series expansion on $0 < |z| < r$, $z = 0$ is not an essential singularity of $f$. $\endgroup$
    – Maxim
    Jan 17 at 20:38
  • $\begingroup$ @Maxim: Right you are! I was considering the completely wrong series. $\endgroup$ Jan 17 at 22:18
  • $\begingroup$ To be clear, the new center of expansion $z = \infty$ is not an essential singularity of $f$ either. $\endgroup$
    – Maxim
    Jan 17 at 22:52

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