Question about Laurent series and dividing the denominator

Question

The problem I am having trouble with is finding the Laurent series for

$f(z)=\frac{1}{z^2\sinh z}$ $0<|z|< \pi$

The example I am looking at says

$\frac{1}{z^2\sinh z}=\frac{1}{z^2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+\dots)}$

and then says$ \frac{1}{z^2\sinh z}=\frac{1}{z^3(1+\frac{z^2}{3!}+\frac{z^4}{5!}+\dots)}$

Then they perform division to find the Laurent series expansion for $f(z),0<|z|<\pi$

My question is why they had to divide the series $\sinh z$ in the denominator by $z$? Wouldn't the answer for the Laurent series expansion be the same if we performed the division without dividing the denominator by $z$? This seems like a needless step but I think for some reason it has to be done. Can someone explain why this step is necessary?

Answer 1

I think the reason they divide by z in the denominator is to get a function free of zeros and poles and one that is entire and so we can replace it with a taylor series. So you can get

$f(z) = \frac{1}{z^3}\sum_{k= 0}^{\infty} a_k z^k = \sum_{k= -3}^{\infty} a_k z^k $

Answer 2

It is a actually a bit of magic with holomorphic functions. The function $f(z)=z/\sinh(z)$ extends to a holomorphic function for $|z|<\pi$ (because $\sinh$ does not vanish here). So it has a power series converging in that disk. Now writing $f(z)=1/(1+u(z))$ we have for e.g. $|z|<0.1$ that $|u(z)|<1$ so we may develop to find the series term by term. For the coefficient to $z^n$ you only need the $n$ first terms. But a power series at 0 is unique so it must be the same as for $f$. Finally for the Laurent series you divide by $z^3$.