Finding a pole when the numerator is 0)

Question

In a more informal way, is it right in saying that $\frac{sin(z)}{z^3}$ has a pole of order 2 at $z=0$ because it would be order 3 (because of the $z^3$) but the $sin(z)$ is also $0$ so we just take one away? Hope that makes sense :/

Thanks!

Answer 1

Correct, $f(z) = \frac{\sin z}{z^3}$ has a pole of order two at $z=0$. Computing the Laurent series gives

$$f(z)= \frac{1}{z^2} - \frac{1}{3!} + \frac{z^2}{5!} - \cdots = \sum_{n=1}^\infty \frac{(-1)^{n+1} z^{(2n+1)-3}}{(2n+1)!}.$$