Z-Transform, Transfer Function, Poles & Zeros I've been working on a question that I'm now stuck on. I need to:

Determine the transfer function and poles-zeros of:

$y[n]=0.5y[n-1]-0.25y[n-2]+x[n]$
So far I've carried out a z-transform in order to get the transfer function (but that's the easy bit)
\begin{equation}
H(Z)={1\over 1-\frac12{z^{-1}}+\frac14{z^{-2}}}.
\end{equation}
However I'm not sure what to do next. The mark scheme basically states:

\begin{equation}{1\over(1-re^{j\theta})(1-re^{-j\theta})}\end{equation} where $r=0.5$ and $\theta=\frac\pi3$

What steps do I need to take to get from the transfer function to the above?
Thanks
 A: HINT:
You need to put z in positive powers:
If you multiply $$H(z)$$ by $$\frac{z^2}{z^2}$$
You will get:
$$ H(z)=\frac{z^2}{z^2-0.5z+0.25}$$
In this form, it is easier to find the poles and zeroes.  You need to factor.  The denominator has two complex poles and numerator two real zeros.
A: \begin{equation}
H(z)=\frac{1}{1-\frac12{z^{-1}}+\frac14{z^{-2}}}
\end{equation}
First pull through a $z^{-2}$ in the denominator and then move it up the the numerator
\begin{equation}
H(z)=\frac{z^2}{z^2-\frac12{z}+\frac14{}}
\end{equation}
For the poles, you have to find the roots of the quadratic $z^2-\frac12{z}+\frac14{}$. We can use the quadratic formula if we like;
$$
z_p = \frac{1}{4}\pm\sqrt{\frac{1}{16}-\frac{1}{4}} = \frac{1}{4}\pm \frac{j\sqrt{3}}{4}=\frac{1}{2}e^{\pm j\pi/3},
$$
where the last equality is conversion to polar form of a complex number.
So you have
\begin{equation}
H(z)=\frac{z^2}{(z-\frac{1}{4}- \frac{j\sqrt{3}}{4})(z-\frac{1}{4}+ \frac{j\sqrt{3}}{4})}=\frac{z^2}{(z-\frac{1}{2}e^{ j\pi/3})(z-\frac{1}{2}e^{- j\pi/3})}
\end{equation}
The denominator shows a complex conjugate pair of poles, and the numerator shows a double-zero at $z=0$.
