I am given the function \begin{equation} f(z) = \dfrac{cos(z)-1}{z^3(z-2)(z-2\pi)} \end{equation} and want to determine the poles. The solution tells me $z_0 = 0$ is a pole of order $3$.
From lectures I know that $z_0$ is a pole of order $m$ if:
\begin{equation} \lim_{z\rightarrow z_0}(z-z_0)^m \cdot f(z) = c ~~~ \text{and} ~~~ c\neq 0 \end{equation} But for $f$: \begin{equation} \lim_{z\rightarrow 0}z^3 \cdot \dfrac{cos(z)-1}{z^3(z-2)(z-2\pi)} = \lim_{z\rightarrow 0}\dfrac{cos(z)-1}{(z-2)(z-2\pi)} = \dfrac{cos(0) - 1}{4\pi} = 0 \end{equation}
How do I know that $z_0 = 0$ is a pole of order $3$ despite the above limit being equal to $0$? Is there an easy way to generally determine poles and their order?
Thanks in advance.
The solution is wrong, it is a pole of order $1$. An easy way to determine this for an expression of the form $\frac{f}{g}$ is to compute the orders of the roots of $f$ and $g$ via derivatives. Since $f(0) = 0$, $f'(0) = 0$ and $f''(0) = -1$, the numerator has a second order root at $0$. Similarly, $g$ has a third order root at $0$ (or in this case just read it off as it is already factorized). Hence two orders 'cancel' and you get a pole of order $3-2=1$.