How to compute magnitude and phase response from transfer function in Z-domain? I have a transfer function $$H(z)=\frac{1+1.2z^{-1}+0.8z^{z^-2}}{1-0.9z^{-1}}$$ from which I'm supposed to sketch the magnitude and phase response. I know that you can transform $z=e^{j\omega}$ to get the frequency response, but I don't really understand what it means. The magnitude response is then $$\left|H(e^{j\omega})\right|=\left|\frac{1+1.2e^{-j\omega}+0.8e^{-j2\omega}}{1-0.9e^{-j\omega}}\right|,$$ but how do I calculate the values from here?

For example, if we set $\omega=0$, then the result is $30$. But if we set $\omega=\pi/4$, then the result is $$\left|\frac{1+1.2e^{-j\pi/4}+0.8e^{-j\pi/2}}{1-0.9e^{-j\pi/4}}\right|.$$ How does one solve this?

I also know that the formula for phase response is $$\arg(H(e^{j\omega}))=\frac{\Im(H(e^{j\omega}))}{\Re(H(e^{j\omega}))}$$ but I'm having a hard time to understand how to use it and what phase response actually is. Can you give me an example of how to calculate the phase response of a certain angle?
Use euler identity to expand the exponential to cos and sin. Then rationalize the denominator, and the last step is to group it to real part and imaginary part. To compute the magnitude just use the pythagoras theorem.

For an example I'll use simple transfer function like: $$H(z) = \frac {z}{z-0.5}$$
Then substitue the $z$:$$H(e^{jw})=\frac{e^{jw}}{e^{jw}-0.5}$$
Expand to cos and sin:
$$H(e^{jw})=\frac{cos(w)+jsin(w)}{cos(w)+jsin(w)-0.5}$$
Rationalize the denominator, group into real and imaginary part:
$$\begin{eqnarray*}H(e^{jw})&amp;=&amp;\frac{cos(w)+jsin(w)}{cos(w)+jsin(w)-0.5}\cdot\frac{cos(w)-jsin(w)-0.5}{cos(w)-jsin(w)-0.5}\\
&amp;=&amp;\frac {e^{jw}\cdot (e^{-jw}-0.5)}{(e^{jw}-0.5)\cdot (e^{-jw}-0.5)}\\
&amp;=&amp;\frac {e^{jw-jw}-0.5e^{jw}}{e^{jw-jw}-0.5e^{jw}-0.5e^{-jw}+0.25}\\
&amp;=&amp;\frac {1-0.5e^{jw}}{1.25-0.5cos(w)-j0.5sin(w)-0.5cos(w)+j0.5sin(w)}\\
&amp;=&amp;\frac {1-0.5cos(w)-j0.5sin(w)}{1.25-cos(w)}\\
&amp;=&amp;\frac {1-0.5cos(w)}{1.25-cos(w)} - j\frac {0.5sin(w)}{1.25-cos(w)}\\
\end{eqnarray*}$$
Compute the magnitude:
$$ \begin{eqnarray*} \left|H(e^{jw})\right| &amp;=&amp; \sqrt{\Re(H(e^{jw}))^2+\Im(H(e^{jw}))^2}\\
&amp;=&amp; \frac {\sqrt{(1-0.5cos(w))^2+(0.5sin(w))^2}} {1.25-cos(w)}\\
&amp;=&amp; \frac {\sqrt{1-cos(w)+0.25cos^2(w)+0.25sin^2(w)}} {1.25-cos(w)}\\
&amp;=&amp; \frac {\sqrt{1-cos(w)+0.25(cos^2(w)+sin^2(w))}} {1.25-cos(w)}\\
&amp;=&amp; \frac {\sqrt{1-cos(w)+0.25}} {1.25-cos(w)}\\
&amp;=&amp; \frac {\sqrt{1.25-cos(w)}} {1.25-cos(w)}\\
\end{eqnarray*}$$
And you're done!