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I know that we should use Laurent series to expand a function around a singularity and Taylor series otherwise. But there is a few aspects that I don´t understand.

Imagine $\sin\left(\frac{1}{z}\right)$. I know $\sin\left(\frac{1}{z}\right)$ has an essential singularity at $z=0$. I don´t understand how do I expand $\sin\left(\frac{1}{z}\right)$ in Laurent series (I know how to expand $\sin(z)$ in Taylor series). Basically I don´t understand the difference between the formula of Laurent and Taylor series. How someone give me some intuition?

Also how could I expand $\sin\left(\frac{1}{z}\right)$ in Laurent series around $z=0$ and how can I tell that it is an essential singularity based on the expansion?

Thanks!

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$ \sin z = \displaystyle \sum_{n=0}^{\infty} \frac{ (-1)^n z^{2n+1}}{(2n+1)!}$ if $|z|< \infty$

Then if $|z| < \infty \rightarrow \displaystyle 0<{|\frac{1}{z}|} < \infty$

$ \sin{\frac{1}{z}} = \displaystyle \sum_{n=0}^{\infty} \frac{ (-1)^n}{z^{2n+1}(2n+1)!}$

And $0$ is a essential singularity because we have infinite terms $b_n$ where $b_n= \displaystyle \frac{1}{2\pi i} \int \frac{f(z)}{z^{-n+1}}$. Basically a point $z_0$ is a essential singularity if the Laurent Series of $f(z)$ in $R_1<|z-z_0|<R_2$ has infinite terms $b_n$ where the Laurent Series is $$\sum_{n=0}^{\infty} a_n(z-z_0)^n +\sum_{n=1}^{\infty} \frac{b_n}{(z-z_0)^n} $$

with $a_n=\frac{1}{2 \pi i} \int \frac{f(z)}{(z-z_0)^{n+1}}$ and $b_n= \frac{1}{2 \pi i}\int \frac{f(z)}{(z-z_0)^{-n+1}}$

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  • $\begingroup$ Thanks for your reply! Just to check if I got it right. The principal part of Larent series is when Z appears on the denominator right? If I have an infinite number of non-zero terms in the principal part that means I got a essential singularity, right? If I got 0 non-zero terms then I got a removable singularity and the residue is 0, right? $\endgroup$ – Daniel Oliveira Nov 3 '17 at 19:06
  • $\begingroup$ Yes, you are right! $\endgroup$ – idk Nov 3 '17 at 19:08
  • $\begingroup$ Just one more question. Is it possible to find the residue of sin(1/z) just by looking at the Laurent expansion? $\endgroup$ – Daniel Oliveira Nov 3 '17 at 19:16
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    $\begingroup$ The residue is the coefficient of the first non-negative power in the Laurent expansion. $\endgroup$ – Faraad Armwood Nov 3 '17 at 19:18
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    $\begingroup$ Yes, for example: $$\sin{\frac{1}{z}} = \displaystyle \sum_{n=0}^{\infty} \frac{ (-1)^n}{z^{2n+1}(2n+1)!}= \displaystyle \frac{1}{z} + \sum_{n=1}^{\infty} \frac{ (-1)^n}{z^{2n+1}(2n+1)!}$$ Then the residue is $1$ $\endgroup$ – idk Nov 3 '17 at 19:19
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You have,

$$ \sin(w) = w - \frac{w^3}{3!}+ \frac{w^5}{5!}- \cdots \Rightarrow \sin\left(\frac{1}{z}\right) = \frac{1}{z} - \frac{1}{3!} \cdot \frac{1}{z^3} + \frac{1}{5!} \cdot \frac{1}{z^5} - \cdots$$

which converges for $|z| \not =0$. To show that $z = 0$ is an essential singularity it suffices to just show that,

$$\lim_{z \to 0} \ z \cdot f(z) \not = 0 \ \ \ \ \ \ \ \lim_{z \to 0} \ z^{n+1} \cdot f(z) \not = 0, \forall n \in \mathbb{N}$$

Both are clear from the expansion.

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    $\begingroup$ Actually, it converges for every $z\neq0$. $\endgroup$ – José Carlos Santos Nov 3 '17 at 18:56
  • $\begingroup$ Thanks for the reply, really appreciate it! I understand that expansion, it is like the Taylor expansion. I don´t understand how you get the principal part of the Laurent series. Is this the principal part because Z is on the denominator and when Z is on the numerator it is called the secondary part (aka Taylor series)? $\endgroup$ – Daniel Oliveira Nov 3 '17 at 18:57
  • $\begingroup$ @JoséCarlosSantos: I'm very sorry for that typo. I was definitely thinking about something else when I wrote that last line. $\endgroup$ – Faraad Armwood Nov 3 '17 at 18:59
  • $\begingroup$ @FaraadArmwood And another thing: perhaps that you could use $|\cdot|$ instead of $\|\cdot\|$. $\endgroup$ – José Carlos Santos Nov 3 '17 at 19:01
  • $\begingroup$ @DanielOliveira: I made an edit. $\endgroup$ – Faraad Armwood Nov 3 '17 at 19:10

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