Finding poles and zeros in the z-domain based on the frequency response I am reading a (pretty good) book about audio signal processing. I would really like to understand the basics to hundred percent, but there is an example I just don't get. It says: "Find the poles and zeros for the following 2nd order system:"
$$
H(z)=\frac{1-1.3435z^{-1}+0.9025z^{-2}}{1-0.45z^{-1}+0.55z^{-2}}
$$
The book writer's own answer starts with the following sentence: "The poles and zeros appear in conjugate pairs because the coefficients of H(z) are real-valued," and then he brings the following equation:
$$
H(z)=\frac{(z-0.95e^{j45^{\circ}})(z-0.95e^{-j45^{\circ}})}{(z-0.7416e^{j72.34^{\circ}})(z-0.7416e^{-j72.34^{\circ}})}
$$
The jump from the first to the second equation is too big for me. 
Could someone please explain the intermediate steps?
Thanks!
 A: Multiply the numerator and denominator by $z^{2}$ to get
$$H(z) = \frac{z^{2}-1.3435z + 0.9025}{z^{2} - 0.45z + 0.55}.$$
Recall that a polynomial with real coefficients has roots which come in conjugate pairs. Apply the quadratic formula to the top and bottom to get for the numerator,
$$r_{1},r_{2} = 0.671751\dots \pm j0.671751\dots $$
and for the denominator,
$$\rho_{1},\rho_{2} = 0.225\pm 0.706665 i.$$
Hence
$$H(z)  = \frac{(z-r_{1})(z-r_{2})}{(z-\rho_{1})(z-\rho_{2})}.$$
Write $r_{1},\rho_{1},r_{2},\rho_{2}$ in polar form, e.g. $\lvert r_{1}\rvert e^{j\theta_{r_{1}}}$ etc, using the fact that the conjugate of a number in polar form is equivalent to taking an additional negative sign in the exponential.
Replace $r_{1},r_{2},\rho_{1},\rho_{2}$ with this polar form to get the desired equation. 
As for how you would convert the roots to polar form, it is not very obvious to me that this is reasonably doable by hand without spending a lot of time on it. Especially with the value of 72.34 degrees in the denominator.
