I've got an exercise on Fourier Transform. I have to show that the following is an all-pass filter: I have to show that $|\hat h(\omega)|=1$ where
$$ \hat h(\omega)=\prod\limits_{k=1}^K\dfrac{\bar a_k - e^{-i\omega}}{1-a_ke^{-i\omega}} $$
with $a_k\in\mathbb{C}$ $\forall k\in\mathbb{N}$. My idea was to bound that quantity from above and from below with some expressions and show that being bounded $1\le |\hat h(\omega)| \le 1$ it is equal to one, but could not find a nice way to obtain this. Any other idea or hint? Thanks in advance.
The answer is much easier to obtain than any bounds. From properties of complex numbers and their magnitudes we get that
$$|\hat{h}(\omega)| = \left|\prod_{k=0}^K \frac{\bar{a}_k-e^{-i\omega}}{1-a_ke^{-i\omega}}\right| = \prod_{k=0}^K|e^{i\omega}|\cdot\left| \frac{\bar{a}_k-e^{-i\omega}}{a_k-e^{i\omega}}\right| = \prod_{k=0}^K 1 \cdot 1 = 1$$
using $|e^{i\omega}| = 1$ and $|\bar{z}| = |z|$ since the numerator is the conjugate of the denominator.