Question

If $a$ is a complex number such that $|a|=1$ and $a{z}^2+z+1=0$ has one purely imaginary root, then $\cos \left(\arg \right(a\left)\right)$ is

• A. $\frac{\sqrt{2}-1}{2}$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $\frac{\sqrt{5}-1}{2}$
• D. $\frac{\sqrt{7}-1}{2}$

Answer 1

The correct option is C $\frac{\sqrt{5}-1}{2}$

$a{z}^2+z+1=0\dots \left(1\right)$
Taking conjugate of both sides,
$\overline{a{z}^2+z+1}=\overline{0}$
$\Rightarrow \overline{a}(\overline{z}{)}^2+\overline{z}+\overline{1}=0$
$\Rightarrow \overline{a}{z}^2-z+1=0\dots \left(2\right)$
$($since $\overline{z}=-z$ as $z$ is purely imaginary$)$
Eliminating $z$ from equations $\left(1\right)$ and $\left(2\right)$, we get
$(\overline{a}-a{)}^2+2(a+\overline{a})=0\dots (3)$
Let,
$a=\cos \theta +i\sin \theta (\because |a|=1)$
From equation $\left(3\right)$ we get,
$(-2i\sin \theta {)}^2+2(2\cos \theta )=0$
$\Rightarrow 4{\cos }^2\theta +4\cos \theta -4=0$
$\Rightarrow {\cos }^2\theta +\cos \theta -1=0$
$\Rightarrow \cos \theta =\frac{-1\pm \sqrt{1+4}}{2}$
$\because \theta =\arg \left(a\right)$
$\therefore \cos \left(\arg \right(a\left)\right)=\frac{\sqrt{5}-1}{2}$