Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

3 Answers

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*}

share|improve this answer
add comment

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

share|improve this answer
add comment

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. $$(z^2 -2c)(z-c) = 0 $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.