Answer
1, 0 and $\frac{1}{2}$
Work Step by Step
$g'(z)=\frac{-1}{(z-1)^{2}}-\frac{-1}{z^{2}}$
$=\frac{1}{z^{2}}-\frac{1}{(z-1)^{2}}=\frac{1-2z}{(z-1)^{2}z^{2}}$
$g'(z)=0\implies 1-2z=0\implies 1=2z\implies z=\frac{1}{2}$
$g'(z)$ does not exist when $z=1$ or when $z=0$ as the denominator vanishes in both of the cases.
The critical points of $g$ are those $z$ values for which $g'(z)=0$ or $g'(z)$ does not exist. Therefore, the critical points are 1, 0 and $\frac{1}{2}$.
