Finding the Laurent series of the function $f(z)=\sin\left(\frac{1}{z}\right)(z - 1)$ when $|z| > 0$.

Question

Define $f(z)=\sin\left(\frac{1}{z}\right)(z - 1)$ for $z\in\mathbb{C}$ such that $|z| > 0$. I have understood that when tasked to find the Laurent series of a function $f$ about a point $z_0$, on some set, we want to represent $f(z) = \sum_{n=-\infty}^{\infty}a_n(z - z_0)^n$. This might be a nitpicky question, but how do we achieve this form for our $f$ when expanding about the origin? By using $\sin(z)$'s Taylor expansion about the origin, we get $f(z) = (z - 1)\sum_{n=0}^\infty \frac{(-1)^n}{(2n + 1)!}z^{-(2n + 1)} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n + 1)!}\left(z^{-2n} - z^{-2n - 1}\right)$, and at least I do not see any way to manipulate this series into the canonical form. So is the aforementioned series the Laurent series of $f$ about the origin, or is there more to the story?

Answer 1

We do not write things like $(z^{−2n}−z^{−2n−1})$ inside the sum symbol because it does not fully tell you (easily) what coefficient goes with each power. I would go somewhat like this:

Know: $$\sin(w)=\sum_{n=0}^{+\infty} (-1)^n \frac{w^{2n+1}}{(2n+1)!} ; \hspace{10 pt} |z|<+\infty$$

substitute $w=1/z$ $$\sin(1/z)=\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n+1}} ; \hspace{10 pt} |z|>0$$

calculate and expand the series of $z\sin(1/z)$: $$z\sin(1/z)=\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n}} =1-\frac{1}{3!z^2}+\frac{1}{5!z^4}-\frac{1}{7!z^6}+\dots; \hspace{10 pt} |z|>0$$

calculate and expand the series of $-\sin(1/z)$: $$-\sin(1/z)=\sum_{n=0}^{+\infty} \frac{(-1)^{n+1}}{(2n+1)!} \frac{1}{z^{2n+1}} =-\frac{1}{z}+\frac{1}{3!z^3}-\frac{1}{5!z^5}+\dots; \hspace{10 pt} |z|>0$$

add both expressions: $$z\sin(1/z)-\sin(1/z) =\left(1-\frac{1}{3!z^2}+\frac{1}{5!z^4}-\frac{1}{7!z^6}+\dots \right)+\left(-\frac{1}{z}+\frac{1}{3!z^3}-\frac{1}{5!z^5}+\dots \right) $$

$$(z-1)\sin(1/z) = 1 -\frac{1}{z} -\frac{1}{3!z^2} +\frac{1}{3!z^3}+\frac{1}{5!z^4}-\frac{1}{5!z^5}-\frac{1}{7!z^6}+\dots \hspace{10 pt} |z|>0$$

And in that last expression it is much clearer to see the regular and principal part of the Laurent expansion, and also much clear to see the pattern involving each power of $z$ and their coefficients.

It doesn't matter if you can't fit the expression into a closed summation form, it is still a Laurent series (never forget the "$\dots$" !).