Let z=eiα and w=eiβ z2+w2=1⇒e2iα+e2iβ=1 ⇒ cos‌2‌α+cos‌2‌β=1 and sin‌2‌α+sin2β=0 ⇒ 2‌cos(α+β)‌cos(α−β)=1 and 2‌sin(α+β)‌cos(α−β)=0 ⇒ sin(α+β)=0⇒α+β=nπ for α+β=0, we have cos‌2‌α=
1
2
⇒4‌pairs(α,β)‌for‌α∈[0,2π) for α+β=π, we have cos‌2‌α=