Figuring magnitude and phase response I've got a linear time-invariant system $$y[n]=\frac{8}{9}y[n-1]+x[n]$$
which I transformed into a transfer function $$Y(z)=\frac{8}{9}Y(z)*z^{-1}+X(z) =&gt;\frac{Y(z)}{X(z)}=\frac{1}{1-\frac{8}{9}*z^{-1}}=\frac{z}{z-\frac{8}{9}}$$
The pole is positive, which would make the system stabile. 
However, I am having some trouble getting the magnitude/phase responses $|H(\omega)|$ and $\theta(\omega)$. 
How should I proceed with the system to figure these out? I've read about using the substitution for $z$ as$$Y(z)=\frac{z}{z-\frac{8}{9}}=\frac{e^{iw}}{e^{iw}-\frac{8}{9}}$$
or even continue to transform the exponentials to their $\cos$ + $\sin$ forms, but i am not sure how i should proceed with them.

As an addition, how would the system change if I would to swap $x[n]$ with a sin or a cosine of some value? I assume it would be the same path towards the responses but with different values.
$$
|H(z)| = \sqrt{\Re^2 H(z) + \Im^2 H(z)} \\
\theta \big(H(z)\big) = \arctan \frac{\Im H(z)}{\Re H(z)} \\
$$

$\Re z$ is the real part of z;

$\Im z$ is imaginary part.

For a single-pole filter $y_n = b_0 x_n - a_1 y_{n-1}$,

$$
H(z) = \frac{b_0}{1 + a_1 z^{-1}}; \\
\Re H(z) = b_0 \frac{1 + a_1 \cos \omega}{1 + 2 a_1 \cos \omega + a_1^2}; \\
\Im H(z) = b_0 \frac{a_1 \sin \omega}{1 + 2 a_1 \cos \omega + a_1^2}.
$$

With $b_0 = 1$ and $a_1 = -\frac{8}{9}$, your filter may be unstable.

For a low-pass filter, try $b_0 = \frac{1}{9}, a_1 = -\frac{8}{9}$.