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Suppose the equation $(E):z^2-2\sin(\alpha)z+2(1+\cos(\alpha))=0$ / $z\in \mathbb{C}$.

I tried to calculate the discriminant but I could determinate it's sign(there is a hint $\Im (z_{1})\ge \Im(z_{2})$ / $z_{2}$ and $z_{1}$ are the two solution of the equation.

In a second case we suppose $(E_{2}):z^3-(1+i\sqrt{3})z^2-2(1+i\sqrt{3})z-4+4i\sqrt{3}$.

In this case I can't use the formula of solving the Cubic polynomial and I know it has two solutions.

So the question is how can I solve those equations?

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up vote 1 down vote accepted

The roots of the first equation are given by \begin{align*}\sin \alpha \pm \sqrt{\sin^2\alpha - 2(1+\cos \alpha)}&= \sin \alpha \pm \sqrt{-\cos^2\alpha - 1 -2 \cos \alpha}\\ &= \sin \alpha \pm \sqrt{-(\cos\alpha + 1)^2} = \sin \alpha \pm (\cos\alpha + 1)i. \end{align*}

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The quadratic formula works just fine, but perhaps you'd have to extract a square root of a complex number.

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the cubic is a rigged question. so, noting that $(1+i\sqrt{3})^2 = -2+\sqrt{3}$ by setting $c=1+i\sqrt{3}$ the equation becomes:

$$ z^3 - cz^2 - 2cz + 2c^2 = 0 $$ i.e. $$(z^2 -2c)(z-c) = 0 $$

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