zeroes of the orthogonal polynomials are simple

Question

I am reading the proof to Theorem 3.2. I am particularly interested in how the roots of the orthogonal polynomials $p_{n+1}$ becomes simple. Please find attached the proof but I am trying to convince why the integral $$0= \int(x-\alpha)^2+ \beta^2 |\frac{p_{n+1}}{(x-\alpha)^2+ \beta^2}|$$ implies that $\beta =0$. please guys, can anyone please explain the implication part to me?

Answer 1

The proof starts off by assuming that $p_{n+1}$ has a pair of complex roots $\alpha \pm i\beta$ and reaches the conclusion that

$$ \int_a^b \left( (x-\alpha)^2 + \beta^2 \right) \left| \frac{p_{n+1}}{(x-\alpha)^2 + \beta^2} \right|^2 \omega(x) \, dx = 0$$

but all three factors are greater than or equal to $0$ for all $x \in (a,b)$ with a finite number of zeros. That implies the integral must be greater than 0, so we have a contradiction and the assumption that $p_{n+1}$ has a pair of complex conjugate zeros must be false.