# Proving $\arg(a)\equiv \alpha\Longleftrightarrow z_1z_2 \in \mathbb{R}$

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

• 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

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