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!
Best Answer
$ \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 $b_n= \frac{1}{2 \pi i}\int \frac{f(z)}{(z-z_0)^{-n+1}}$