Order of pole of $f(z) = \frac{\cos z-1}{z^3}$.

Question

I am looking at a practice problem for singularities of complex functions, namely $f(z) = \frac{\cos z-1}{z^3}$. According to Wolfram alpha, it is a pole of order one, but when evaluating $\lim_{z\to 0}\frac{\cos z-1}{z^3}(z-0)^k$ I get $\frac{-\sin z}{-k(3-k)z^{2-k}}$, which only exists for $k = 2$, indicating that the pole is of order $2$. Does anyone know where I'm making a mistake?

Answer 1

It looks like you are trying to apply L'Hopital's rule. But you'll find it easier simply to use the power series for cosine:

$$\frac{\cos z - 1}{z^3} = \frac{-z^2/2! + z^4/4! - \cdots}{z^3} $$

You then have a Laurent expansion and can see the poles and calculate the residue very easily.

Answer 2

I figured out that of I'd have to differentiate again, because of course $\lim_{z\to 0}\frac{-\sin z}{-k(3-k)z^{2-k}}$ will be zero again, so that I will have to evaluate $\lim_{z\to 0}\frac{\cos z}{(2-k)(3-k)z^{1-k}}$ which will yield the expected result.

Answer 3

$$\frac{\cos z-1}{z^3}=\frac{1-z^2/2+z^4/24+...-1}{z^3}=-\frac{1}{2z}+z/24+...$$ So there is a ploe of order 1 at $z=0$.