Laurent series of $\frac{1-\cos(z)}{z^2}$ How do I calculate the Laurent series of $\frac{1-\cos(z)}{z^2}$? (I know the general formula as is shown here )
 A: Hint: $$\cos{z} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - ...$$
so it follows that
$$1 - \cos{z} = \frac{z^2}{2!} - \frac{z^4}{4!} + \frac{z^6}{6!} = z^2 \left(\frac{1}{2!} - \frac{z^2}{4!} + ...\right)$$
A: If you don't want to use the general formula, but you know the series for $\cos z$, then you can use them.
I understand that $f(z)=\frac{1-\cos z}{z^2}$ is not defined at $z=0$, so the function should be competed as
$$f(0)=\lim_{z\to0}\frac{1-\cos z}{z^2}\quad\text{if exists}$$
which makes it difficult for a direct approach.  In this case the limit exists and is $1/2$.

update (to show the general formula to get the Laurent series)
Let's assume that in a neighborhood of $z=c$, then \begin{align}f(z)&=a_0+a_1(z-c)+a_2(z-c)^2+\ldots\\&=\sum_{k=0}^\infty a_k(z-c)^k\end{align}
If we derivate $n$ times we have:\begin{align}f'(z)&=\frac{d}{dz}\sum_{k=0}^\infty a_k(z-c)^k\\&=\sum_{k=0}^\infty ka_k(z-c)^{k-1}\\&=\sum_{k=1}^\infty ka_k(z-x)^{k-1}\\&=\sum_{k=0}^\infty(k+1)a_{k+1}(z-c)^k\end{align}
Evaluated at $z=c$ then $f'(c)=a_1$.
Repeat the process $n$ times: \begin{align}f''(z)&=\sum_{k=0}^\infty(k+1)(k+2)a_{k+2}(z-c)^k\\f'''(z)&=\sum_{k=0}^\infty(k+1)(k+2)(k+3)a_{k+k}(z-c)^k\\f^{(n)}(z)&=\sum_{k=0}^\infty\frac{(k+n)!}{k!}a_{k+n}(z-c)^k\end{align}
Evaluate at $z=c$ and $f''(c)=2!a_2,f'''(c)=3!a_3\ldots f^{(n)}(c)=n!a_n$.
So you can get $a_n=\frac1{n!}f^{(n)}(z)|_{z=c}$.
