Equality between 2-Norm of polynomials Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+ \dots a_0$ be a polynomial of $\mathbb{C}[x]$, we define 
$$||f||_2= \sqrt{ \sum_i|a_i|^2}$$
Now, studying Computational Algebra, I've read about the following equality:
$$||q(x)(x-\alpha)||_2=||q(x)(\bar{\alpha}x-1)||_2$$
where $q(x) \in \mathbb{C}[x]$ and $\alpha \in \mathbb{C}$.
I think induction on degree of $q$, in some way, works to prove it, but I've tried with no results...
Any ideas to proof of this fact?
Thank you

If we let $q(x) = \sum_{i=0}^n a_i x^i$ then, if we define $a_{-1}=a_{n+1}=0$,  we have
$$q(x)(x-\alpha) = \sum_{i=0}^{n+1} (a_{i-1}-\alpha a_i) x^i$$ 
$$q(x)(\overline{\alpha}x-1) = \sum_{i=0}^{n+1} (\overline{\alpha}a_{i-1}-a_i) x^i$$
From the definition of the two norm being equal we get the equation
$$\sum_{i=0}^{n+1} |a_{i-1}-\alpha a_i|^2 = \sum_{i=0}^{n+1} |\overline{\alpha}a_{i-1}-a_i|^2$$
Expand the square using $|x|^2 = x \overline{x}$ on both sides to get (after cancellation)
$$\sum_{i=0}^{n+1}(|\alpha|^2-1)(|a_{i-1}|^2 - |a_i|^2) = 0$$
Finally, after a change of the summation index (and use of $a_{-1} = a_{n+1} = 0$) we find that the left hand side evaluates to $0$ so the two norms are indeed equal.
