How do you solve the equation $ (z^2-1)^2 = 4 ? $ $ (z^2-1)^2 = 4  \iff $$z_1 = 3 $ and $ z_2=-1$
$arg(z_1)= 0 ,  arg(z_2) = \pi$
$$ z_1 = \sqrt{3} \left(\cos\left(\frac{2\pi k}{2}\right) + i\sin\left(\frac{2\pi k}{2}\right)\right)$$
$$z_2 = i \left(\cos\left(\frac{2\pi k}{2} +\pi\right) + i\sin\left(\frac{2\pi k}{2} +\pi\right)\right) \mid k= \{ 0,1 \}$$
$$ z = \{ \sqrt{3}, -\sqrt{3}, i, -i \}$$
Is this correct?

$(z^2-1)^2=4\iff$
$z^2-1=2,-2\iff$
$z^2=3,-1\iff$
$z=\sqrt{3},-\sqrt{3},\sqrt{-1},-\sqrt{-1}\iff$
$z=\sqrt{3},-\sqrt{3},i,-i$
