I want to calculate : $$ \int_0^{2\pi} \dfrac{\mathrm{d} \theta}{2+\cos(\theta)} $$
I use $z=\mathrm{e}^{\mathrm{i} \theta}$ and residue theorem : $$\int_0^{2\pi} \dfrac{\mathrm{d} \theta}{2+\cos(\theta)}=\int_\gamma \dfrac{\mathrm{d}z}{\mathrm{i} z\left(2+\frac{z+z^{-1}}{2}\right)}=\dfrac{2}{i}\int_\gamma \dfrac{\mathrm{d} z}{z^2+4z+1} = 2 \pi \mathrm{Res}\left(f(z),\alpha\right)$$
Where $f(z)=\dfrac{1}{z^2+4z+1}$ and $\alpha=\sqrt{3}-2$. Using limits method, I found : $$\mathrm{Res}\left(f(z),\alpha\right)=\lim_{z\to \alpha} (z-\alpha)f(z)=\lim_{z\to \alpha} \dfrac{1}{z+2+\sqrt 3}= \dfrac{1}{2\sqrt{3}} $$
And here comes my problem. I try to compute the Laurent serie of $f$ in $\alpha$ to found the residue but unfortunately I miss something. With $\beta=-2-\sqrt 3$ the other root and putting $t=z-\alpha$, I found :
$$f(z)=\dfrac{1}{(z-\alpha)(z-\beta)}=\dfrac{1}{t(t-4)}=-\dfrac{1}{4t}\sum_{n\geq 0} \left(\dfrac{t}{4}\right)^n=-\sum_{n\geq -1} \dfrac{t^n}{4^{n+2}}$$
And so, by definition, $\mathrm{Res}\left(f(z),\alpha\right)$ should be $-\frac{1}{4}$ which is wrong...
Someone could show me the mistake and explain how to find the Laurent serie ?
$f(z)=\dfrac{1}{(z-\alpha)(z-\beta)}\\ t = z-\alpha\\ f(t)=\dfrac{1}{t(t+\alpha - \beta)} =\dfrac{1}{t(t + 2\sqrt3)} $