Integrals of trignometric functions Question is to Prove that :
$$\int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\frac{2\pi}{\sqrt{a^2-b^2}} \text{for}\  a>b>0$$
using residue theory.
What i have done so far is :
I transformed functio of $\theta$ as function of complex entity $z$ with $z=e^{i\theta}$.
Then, we have : $dz=i.e^{i\theta}.d\theta\Rightarrow dz=i.z.d\theta \Rightarrow d\theta =\frac{dz}{iz}$ 
$$\Rightarrow \int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\int_{|z|\leq 1} \frac{1}{a+b.\frac{1}{2i}.(\frac{z^2-1}{z})}.\frac{dz}{iz}$$
$$\int_{|z|\leq 1} \frac{1}{a+b.\frac{1}{2i}.(\frac{z^2-1}{z})}.\frac{dz}{iz}= 2 \int_{|z|\leq 1}  \frac{1}{bz^2+(2ai)z-b} dz$$
So, now the actual problem comes :
I have to find poles for this and use residue theorem to evaluate the integrals :
$bz^2+(2ai)z-b=0 \Rightarrow z=\frac{-2ai \pm \sqrt{-4a^2+4b^2}}{2b}$. But, $a>b$ so, i would like to see this as
$\frac{-2ai \pm 2i\sqrt{a^2-b^2}}{2b}=\frac{-ai \pm  i \sqrt{a^2-b^2}}{b}$
Now, i am not able to decide which of $\frac{-ai +  i \sqrt{a^2-b^2}}{b}$ and $\frac{-ai -  i \sqrt{a^2-b^2}}{b}$ is in $\{ z\in \mathbb{C} : |z|\leq 1\}$
I tried considering their modulus.. 
Please help me to understand this problem.
As this is just my third problem in "solving integrals using residue theory", I do not have much experience to tackle this problem...
Once i know what are the poles in $\{ z\in \mathbb{C} : |z|\leq 1\}$ then i know how to use residue theorem and conclude required result.
Thank You.
EDIT : 
we need to consider $|\frac{-ai -  i \sqrt{a^2-b^2}}{b}|=\big|-i(\frac{a+\sqrt{a^2-b^2}}{b})\big|=\big|\frac{a+\sqrt{a^2-b^2}}{b}\big| \leq |\frac{a}{b}|+|\sqrt{(\frac{a}{b})^2-1}|$
As $a>b$, we have $\frac{a}{b}>1$ SO, I do not have specific upper bound for $\frac{a}{b}$ ..  I am confused..
even if i have $|\frac{-ai +  i \sqrt{a^2-b^2}}{b}|$ then also i can not see..

Consider your roots:
$$z_{\pm} = -i \left [ \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b}\right)^2-1}\right]$$
Note that, when $a>b$, $a/b > 1$.  Therefore $|z_+| > 1$.  Note also that
$$|z_-| = \frac{a}{b} - \sqrt{\left ( \frac{a}{b}\right)^2-1} = \frac{1}{\frac{a}{b} + \sqrt{\left ( \frac{a}{b}\right)^2-1}} = \frac{1}{|z_+|} < 1$$
The residue at the pole $z=z_-$ is
$$\frac{2}{2 b z_-+i 2 a}  = \frac{i}{a-\sqrt{a^2-b^2}-a} = \frac{-i}{\sqrt{a^2-b^2}}$$
The value of the integral is then $2 \pi/\sqrt{a^2-b^2}$, as was to be demonstrated.
