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This is a probably basic question about Laurent series. Say $g(z)$ is an analytic function, that $g(0) = 0$, and $f(z) = g(z)/z$. My textbook says $z = 0$ is a removable singularity of $f(z)$.

A removable singularity is defined as a singularity where the principal part of the Laurent series is zero. The Laurent series is:

$$\sum_{k=-\infty}^\infty c_k (z-z_0)^{k}$$

$$c_k = \frac{1}{2\pi i} \oint_C \frac{f(\xi)}{(\xi - z_0)^{k+1}} d\xi$$

The "principal part" of the Laurent series refers to the terms where $k \le -1$. Now if I look at the terms $k=-2, -1, 0$ of the Laurent series of $f(z)$ around $z = 0$:

$$c_{-2} = \frac{1}{2\pi i} \oint_C g(\xi) d\xi = 0$$

(This follows from Cauchy's integral theorem and the fact that $g$ is analytic.)

$$c_{-1} = \frac{1}{2\pi i} \oint_C \frac{g(\xi)}{\xi} d\xi$$

$$c_{0} = \frac{1}{2\pi i} \oint_C \frac{g(\xi)}{\xi^2} d\xi$$

Now it seems to me that $c_{-1}$ is not zero. Therefore, the principal part is not zero, and $z=0$ is a pole of order one.

Any hint or pointer as to where my reasoning goes wrong would be welcome.

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  • $\begingroup$ If $g$ is analytic (in $0$), then $f(z) = g(z)/z$ has a removable singularity in $0$ if and only if $g(0) = 0$. $\endgroup$ – Daniel Fischer Oct 29 '13 at 15:40
  • $\begingroup$ @DanielFischer: It is given that $g(0) = 0$. How do you apply that fact to make $z=0$ a removable singularity? $\endgroup$ – Andomar Oct 29 '13 at 15:50
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    $\begingroup$ If $g(0) = 0$, you have $\lim\limits_{z\to 0} \frac{g(z)}{z} = g'(0)$, so $f$ is in particular bounded in a punctured neighbourhood of $0$, and that means the singularity is removable. $\endgroup$ – Daniel Fischer Oct 29 '13 at 15:53
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    $\begingroup$ "A removable singularity is defined as a singularity where the principal part of the Laurent series is zero." Yuck! What book is that? Anyway, it also works with that definition well enough. $g$ being analytic means the principal part of the Laurent series of $g$ is zero. $g(0) = 0$ means $a_0 = 0$. So the Laurent series of $f$ is $\sum\limits_{n=0}^\infty a_{n+1}z^n$ if $\sum\limits_{n=0}^\infty a_nz^n$ is the Laurent series of $g$. $\endgroup$ – Daniel Fischer Oct 29 '13 at 15:56
  • $\begingroup$ @AWertheim, DanielFisher: Awesome, I get it now! Thanks a lot. $\endgroup$ – Andomar Oct 29 '13 at 16:00
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There must be some other assumptions on the function $g(z)$, otherwise there's no reason a priori to expect that $z = 0$ is a removable singularity of $g(z)/z$.

For instance, $g(z) = 1$ is analytic, but $f(z) = 1/z$ has simple pole (not a removable singularity) at $z = 0$.

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  • $\begingroup$ It's also given that $g(0) = 0$ (I'll add it to the question.) $\endgroup$ – Andomar Oct 29 '13 at 15:51

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