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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?

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  • $\begingroup$ It mean that $\beta \in \{|z|<1\}$ whereas $\alpha \notin \{| z|\leq 1\}$. $\endgroup$ – Walace May 24 '20 at 11:28
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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 $|\beta|>1$.

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