Solve $\int_{-\pi}^{\pi}{\frac{d\phi}{a+b\cos \phi}}$ using residues.

Question

I do a variable substitution by adding $\pi$ to $\phi$ and the sign of $\cos$ changes. $\int_0^{2\pi}{\dfrac{d\phi}{a-b\cos \phi}} = 2\pi \sum_{k=1}^n {res \frac{1}{z} \big [(a-b \frac{1}{2}(z + \frac{1}{z}))^{-1}, z_k \big]}$ where $z_1,...,z_n$ are the function's those poles that have a norm less than $1$ ($|z| < 1$). Now, if I simplify the function I obtain $\dfrac{-2}{bz^2-2az+b}$ and its poles can be obtained by finding the roots of the polynomial which are $z_{12} = \dfrac{a \pm \sqrt{a^2-b^2}}{b}$. Finally, the derivative of the polynomial is $2bz - 2a$ and on $z_{12}$ is not $0$ so we can use a formula (which says that the residue is equal to the value of a function on z obtained by leaving the nominator as it is and changing the denominator with the derivative of it), Thus the residues are equal to $\dfrac{-1}{\pm \sqrt{a^2+b^2}}$. But, I have made a mistake somewhere, because then the some of these will yield $0$. Also, the book says the answer is $\dfrac{2 \pi}{\sqrt{a^2+b^2}}$.