If $z = \frac{1}{2}$, $1 = -\frac{1}{2}A$, so $A = -2$.
$\frac{z}{2z^2 - 3z + 1} = \frac{1}{2} \left( \frac{-2}{2z - 1} + \frac{2}{z - 1} \right) = \frac{1}{2} \left( \frac{-1}{z - \frac{1}{2}} + \frac{2}{z - 1} \right) = \frac{1}{2} \left( \frac{1}{\frac{1}{2} -
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