I have a question about a discrete time filter. All I have is the pole-zero plot and I have to calculate the impulse, phase and magnitude response. To make this a proper fraction I used polynomial division:
$$H(z) = \frac{z-0.5}{z+0.5}= 1-\frac{1}{z+0.5}$$
Now I bracket out the z, so I can use the z-Transform $$H(z) = 1-z^{-1} \cdot \frac{1}{1+0.5*z^{-1}}$$
When I use the transformation, I get the impulse response $h[n]= \delta[n] - (0.5)^{n} \cdot u[n-1]$ Is this correct?
For the phase response I set $e^{j\omega}$ into $H(z)$:
$$H(e^{j\omega}) = 1-\frac{1}{e^{j\omega }+ 0.5}$$
How can I get $|H(e^{j \omega})|$?
You can write
$$ \frac{Y(z)}{U(z)} = \frac{1 - 0.5 z^{-1}}{1 + 0.5 z^{-1}} $$
to obtain the difference equation
$$ h[n] = -0.5 h[n-1] + \delta[n] - 0.5 \delta[n-1] $$
for the impulse response. Assuming $h[-1] = 0$ you can write a few values and figure out the general rule for impulse response instead of calculating the convolution. So,
$$h[0] = 1, h[1] = -1, h[2] = 0.5, h[3] = -(0.5)^2, h[4] = (0.5)^3, \dots$$
The general rule is
$$ h[n] = \delta[n] - (-0.5)^{n-1} u[n-1] $$
where $u$ is the step function.
For the phase response you can use the Euler's identity as @Ian stated in the comments.