$1 + {\alpha}^2 + 2 \alpha \cos(\omega)$ as a square of distance

Question

How to prove that

$$ 1 + {\alpha}^2 + 2 \alpha \cos(\omega) = |1 - \alpha e^{-j\omega}|^2$$

starting from

$$ 1 + {\alpha}^2 + 2 \alpha \cos(\omega)$$

(That is, no backwards proofs of multiplying out $ |1 - \alpha e^{-j\omega}|^2$ to see that it matches $1 + {\alpha}^2 + 2 \alpha \cos(\omega)$.)

Answer 1

Hint

Use $$\alpha^2=\alpha^2(\cos^2\omega+\sin^2\omega)$$and substitute.

Answer 2

assuming $\alpha$ is a real number (not complex):

$$1-2 \alpha cos(\omega)+|\alpha|^2 $$ $$1-2 \alpha \frac{1}{2} (e^{j\omega}+e^{-j\omega})+|\alpha|^2$$ $$1-\alpha e^{j\omega} - \alpha e^{-j\omega} + \alpha^2 e^{j\omega}e^{-j\omega}$$ $$(1-\alpha e^{j\omega}) - \alpha e^{-j\omega}(1-\alpha e^{j\omega})$$ $$(1-\alpha e^{j\omega})(1-\alpha e^{-j\omega})$$ $$|1-\alpha e^{j\omega}|^2$$