How do you calculate phase of $\text{DTFT}$ when exponential is in denominator?

Question

$$Xe^{jw}=\frac{1}{1-ae^{jw}}$$

How is the phase of this derived as $\tan ^{-1}\left(a*\sqrt{\left(1-a^2\right)}\right)$?

Answer 1

$$\begin{align*} X\left(e^{j\omega}\right) &= \dfrac{1}{1-ae^{-j\omega}}\\ \\ &= \dfrac{1}{1-ae^{-j\omega}}\cdot\dfrac{1-ae^{j\omega}}{1-ae^{j\omega}} \\ \\ &= \dfrac{1-ae^{j\omega}}{1 -ae^{-j\omega} -ae^{j\omega} + a^2}\\ \\ &= \dfrac{1-ae^{j\omega}}{1 -2a\cos\omega + a^2}\\ \\ &= \dfrac{1-a\cos\omega}{1 -2a\cos\omega + a^2}+ j\dfrac{-a\sin\omega}{1 -2a\cos\omega + a^2}\\ \end{align*}$$

So the phase is given by:

$$\begin{align*}\angle X\left(e^{j\omega}\right) &= \mathrm{arctan2}\left(\dfrac{-a\sin\omega}{1 -2a\cos\omega + a^2}, \dfrac{1-a\cos\omega}{1 -2a\cos\omega + a^2}\right)\\ \end{align*}$$

And the magnitude is given by:

$$\begin{align*}\left|X\left(e^{j\omega}\right)\right| &= \dfrac{\sqrt{(1-a\cos\omega)^2+(-a\sin\omega)^2}}{1 -2a\cos\omega + a^2}\\ \\ &= \dfrac{\sqrt{1-2a\cos\omega+a^2\cos^2\omega+a^2\sin^2\omega}}{1 -2a\cos\omega + a^2}\\ \\ &= \dfrac{1}{\sqrt{1 -2a\cos\omega + a^2}}\\ \end{align*}$$