Computing the Laurent series of $\frac{1-z^4}{z^4}$ at $z_0=0$

Question

I want to find the Laurent series for $$\frac{1-z^4}{z^4}$$ at $z_0=0$

I would write $$ \frac{1-z^4}{z^4} = \frac{1}{z^4} \frac{1}{\frac{1}{1-z^4}} = \frac{1}{\sum_{k=1}^{\infty} z^{4k}}= a_1z^4 + a_2 z^8 +\ldots \implies a_i=1 $$

Is that right?

Answer 1

No, that is not right. Just note that$$\frac{1-z^4}{z^4}=z^{-4}-1.$$That's all: the Laurent series that you're after is $z^{-4}-1$.