# Computing Definite Integrals using Complex Variables and Contour Integration

I was trying to prove the definite integral $$\displaystyle{\int_{0}^{2\pi}}\frac{d\theta}{a-b\cos{\theta}}=\frac{2\pi}{\sqrt{a^2-b^2}}\quad(a>b\geq0)$$ by using contour integration and the Residue Theorem. I checked the solution and there I am having a doubt. It says :

Let $$z = e^{i\theta}$$. Then $$\cos {\theta} = \frac{1}{2}(z + z^{−1})$$. It is obvious that the integration over $$\theta$$ from $$0$$ to $$2\pi$$ now becomes an integral over $$z$$ around the unit circle in the complex plane, traversed once in the positive sense. Since $$dz = e^{iθ} i dθ$$, or $$dθ = dz/(iz)$$, we get, $$\displaystyle{\int_{0}^{2\pi}}\frac{d\theta}{a-b\cos{\theta}}=\frac{2i}{b}\oint_{|z|=1}\frac{dz}{(z-\alpha)(z-\beta)}$$ where $$\alpha$$ and $$\beta$$ are the roots of $$z^2 − (2a/b)z + 1 = 0$$, i.e., $$\alpha, \beta = (a ± \sqrt{a^2 − b^2)}/b$$ . It is easily checked that the pole at $$\beta$$ lies inside the unit circle, while the pole at $$\alpha$$ lies outside it.

I do not understand the last line (in Bold). How can I check that?

• It mean that $\beta \in \{|z|<1\}$ whereas $\alpha \notin \{| z|\leq 1\}$. May 24, 2020 at 11:28

## 1 Answer

Let $$\alpha=\frac{a-\sqrt{a^2-b^2}}b$$ and $$\beta=\frac{a+\sqrt{a^2-b^2}}b$$. Then $$\alpha\beta=1$$. Besides, $$\beta>0$$ and, since $$\alpha=\frac1\beta$$, $$\alpha>0$$. But then $$0<\alpha<1<\beta$$. So, $$|\alpha|<1$$, and $$|\beta|>1$$.