Find maximum of $|f(z)| = |\frac{(z-1)^2(z+1)}{z^2}|$ on $|z|=\sqrt2, z\in\mathbb{C}$ 
Find maximum of $|f(z)| = |\frac{(z-1)^2(z+1)}{z^2}|$ on $|z|=\sqrt2, z\in\mathbb{C}$

I tried to solve it by parametrization $z = \sqrt{2}(cos(t)+isin(t))$ but it seems too complicated to figure out by differentiation in real line(with variable $t$). Is there another way of figuring out?
note: I know so far that two saddle points of $f(z)$ lie on $|z|=\sqrt2$ but that's all I know.

Note that $f(z) \ne 0$ on $|z|=\sqrt 2$, so representing $f(re^{i\theta})=Re^{i\Phi}$ the local extrema of $|f|=R$ for $r$ constant are given by $\frac{\partial R}{\partial \theta}=0$. 
Note that locally where $f \ne 0$ we have $\log f =\log R +i\Phi$, so using $\frac{\partial \log f}{\partial \theta}=i\frac{zf'}{f}$ we have $\frac{zf'}{f}=\frac{\partial \Phi}{\partial \theta}-i\frac{1}{R}\frac{\partial R}{\partial \theta}$ so we need $\Im \frac{zf'}{f}=0$. 
(the two critical points $\frac{-1 \pm i\sqrt 7}{2}$ which are on the given circle automatically satisfy that)
Since $f(z)=\frac{(z^2-1)(z-1)}{z^2}=z-1-\frac{1}{z}+\frac{1}{z^2}$ we have $zf'=z+\frac{1}{z}-\frac{2}{z^2}=\frac{z^3+z-2}{z^2}=\frac{(z-1)(z^2+z+2)}{z^2}$, so 
$\frac{zf'}{f}=\frac{z^2+z+2}{z^2-1}=1+\frac{z+3}{z^2-1}$, so $\Im \frac{zf'}{f}=0$ implies $\Im ((z+3)(\bar z^2-1))=0$ or (remembering that $|z|^2=2$), $\Im (2\bar z-z+3\bar z^2)=0$ or $(2\bar z-z+3\bar z^2)-(2z-\bar z+3z^2)=0$ which means $(\bar z-z)(\bar z+z+1)=0$ so either we get the two critical points which are precisely the roots of $\bar z+z+1$ or $z=\bar z$ so $z=\pm \sqrt 2$
By inspection $f(\pm \sqrt 2)=\frac{\pm \sqrt 2 -1}{2}$ so the absolute values are $\frac{\sqrt 2 +1}{2}, \frac{\sqrt 2 -1}{2}$, while at the critical points we get $\frac{5}{2}$ hence that is the global maximum. 
