Second Order Transfer Function : Imaginary component in $R$? I have been looking at the following transfer function:

$$ H(z) = \tfrac12 -  \tfrac12 z^{-2} $$

Given the usual method for finding $\theta$ and $R$ in the complex plane, I calculate that $\theta = \frac{\pi}{2}$ and $R = j\sqrt{\frac12}$

This would initially make me think that the conjugate zero pair here is 90° from the real axis, however this imaginary $R$ component has thrown me completely - the equivalent frequency response also seems odd to me as it shows zero at DC and Nyquist with a gentle curve upwards in between.

What is going on here? I think that $R$ in the complex plane can't be imaginary, but don't understand how to arrive at the correct angle and magnitude for the conjugate poles.
HINT:

If you're looking for the zeros of $H(z)$, try $z_0=1$ or $z_0=-1$. If that works, try to figure out why, and derive it yourself. Then, think about what the magnitude and the phase of those zeros might be.

EDIT (in reaction to your comment): If you have a second-order polynomial with real-valued coefficients, then two complex conjugated zeros are not the only option; you can also have two real-valued zeros.