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 complex equation $iz^2+(2-i)az-(1+i)a^2=0$ as $a\in \mathbb{C}^{*}$.

$z_1$ and $z_2$ are the solution of this equation and we have also $z_1*z_2 = a^2(i-1)$.

How can I prove that $\arg(a)\equiv \dfrac{-3\pi}{8}[\dfrac{\pi}{2}]\Longleftrightarrow z_1z_2 \in \mathbb{R} $ ?

share|improve this question
    
What is the relation in question? –  Pedro Tamaroff Feb 27 '13 at 9:32
    
ok I find the solution. –  pourjour Feb 27 '13 at 9:39
    
You're considering equivalence classes, but I fail to see what is the equivalence relation you're considering. Is it $\mod 2\pi$? –  Pedro Tamaroff Feb 27 '13 at 9:40
    
@PeterTamaroff yes –  pourjour Feb 27 '13 at 9:48
add comment

1 Answer 1

up vote 1 down vote accepted

We have

$\begin{eqnarray} z_1z_2 \in \mathbb{R} &\Longleftrightarrow& \arg(a^2(i-1))\equiv 0[\pi] \\ &\Longleftrightarrow& 2\arg(a)+arg(i-1)\equiv 0[\pi] \\ &\Longleftrightarrow& 2\arg(a)+\dfrac{3\pi}{4} \equiv 0 [\pi] \\ &\Longleftrightarrow& \arg(a) \equiv \dfrac{-3\pi}{8}[\dfrac{\pi}{2}] \end{eqnarray}$

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.