Letting software do all the grubby work
$|z^2 + 2z - 3|^2 = x^4+4 x^3+2 x^2 \left(y^2-1\right)+4 x \left(y^2-3\right)+y^4+10 y^2+9$
with $z = \cos(\theta + i \sin(\theta)$ this becomes (after a ton of computer aided simpilfication)
$|z^2 + 2z - 3|^2 = 8 \sin ^2\left(\frac{\theta }{2}\right) (3 \cos (\theta )+5)$
$\dfrac{d}{d\theta} \left( 8 \sin ^2\left(\frac{\theta }{2}\right) (3 \cos (\theta )+5) \right)=8 \sin \left(\frac{\theta }{2}\right) \left(5 \cos \left(\frac{\theta }{2}\right)+3 \cos \left(\frac{3 \theta }{2}\right)\right)$
setting this equal to 0 and solving for $\theta$ we get
$\theta = 0, ~\pi, ~\pi - \tan^{-1}(2\sqrt{2}), -\pi - \tan^{-1}(2\sqrt{2})$
I leave it to you to determine which of these are maxima and which are local minima using the 2nd derivative.