How to prove that a root $z_2=\frac{-a-\sqrt{a^2+b^2}}{2b} <1$ I want to prove that that the integral $I= \int_{0}^{2π} \frac{1}{a+b \cos\theta}\,d \theta$, where $a>0$, $b>0$ and $a>b$, is equal to $ I=\dfrac{2πb}{\sqrt{a^2+b^2}}$ with the method of residues.
My approach is:  
Let $z=e^{i\theta}$ be the parametrization of the circle; I will convert the complex integral to a complex integral on the circle $\gamma : \lvert z \lvert=1$ 
We have $\dfrac{dz}{d\theta}= ie^{i\theta}\Rightarrow dz=iz\,d \theta \Rightarrow \frac{1}{iz}\,dz=d\theta$
Also $\cos\theta = \dfrac{e^{i\theta}+e^{-i\theta}}{z} =\dfrac{z+ \bar{z}}{2}= \dfrac{z(z+ \bar{z})}{2z}=\dfrac{z^2+1}{2z}$
If I replace this quantity in the integral $Ι$, with easy calculations I have:
$$I = \int_{0}^{2π} \frac{2}{bz^2+2az+1}\,dz$$
So, with the factorization of denominator we have $\displaystyle I = \int_{0}^{2π} \frac{2}{(z-z_1)(z-z_2)}\,dz$ where $z_1=\dfrac{-a+\sqrt{a^2+b^2}}{2b}$ and  $z_2=\dfrac{-a-\sqrt{a^2+b^2}}{2b}$ 
The discriminant is positive since we have $a>0$ and $b>0$ . 
To continue with the method of residues and finally find the result, clearly I have to cancel a root from $z_1$ and $z_2$ which is not contained in the circle $\lvert z \lvert=1$.
Intuitively this root is $z_2$. How can I prove that this root is outside this circle? Any obvious condition, I think doesn't show anything. 

Without loss of generality, we can divide everything by $b$, and so look at $c=a/b$:
$$ -\frac{a/b}{2} \pm \frac{\sqrt{(a/b)^2+1}}{2} = -\frac{c}{2} \pm \frac{\sqrt{1+c^2}}{2}. $$
The conditions give $c>1$. Now, $\sqrt{1+c^2}>\sqrt{c^2}=c$, so
$$ -\frac{c}{2} - \frac{\sqrt{1+c^2}}{2} < -\frac{c}{2}-\frac{c}{2} = -c < -1. $$
Meanwhile, to check the other one is inside the circle,
$$ \frac{\sqrt{1+c^2}-c}{2} = \frac{(\sqrt{1+c^2}-c)(\sqrt{1+c^2}+c)}{2(\sqrt{1+c^2}+c)} = \frac{1}{2(\sqrt{1+c^2}+c)}, $$
and this is obviously positive, but also $\sqrt{1+c^2}+c >1+c$, and so
$$ \frac{1}{2(\sqrt{1+c^2}+c)}<\frac{1}{2(1+c)}<1, $$
certainly.
