Question

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

• A. $co{s}^{-1}\left(\frac{\sqrt{5}+1}{2}\right)$
• B. $co{s}^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
• C. $si{n}^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
• D. $si{n}^{-1}\left(\frac{\sqrt{5}+1}{2}\right)$

Answer 1

The correct option is B $co{s}^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$

${az}^2+z+1=0$ (i)
Where $z$ is purely imaginary. Taking conjugate of both sides, $\overline{{az}^2+z+1}=\overline{0}$
$\Rightarrow \overline{a}{\left(\overline{z}\right)}^2+\overline{z}+\overline{1}=0$
$\Rightarrow \overline{a}{\left(z\right)}^2-z+1=0$ (since $\overline{z}=-z$ as $z$ is purely imaginary) ...(ii)
Eliminating $z$ using the equations,(i) and (ii)
$\Rightarrow {(\overline{a}-a)}^2+\overline{z}+\overline{1}=0$
Let $a=\cos \theta +i\sin \theta$ (since ) $\left|a\right|=1$
So ${(-2i\sin \theta )}^2+2\left(2\cos \theta \right)=0$
$\Rightarrow -4{\sin }^2\theta +4\cos \theta =0\Rightarrow {\cos }^2\theta +\cos \theta -1=0$
$\Rightarrow \cos \theta =\frac{-1\pm \sqrt{1+4}}{2}$
only feasible value of $\cos \theta$ is $\frac{\sqrt{5}-1}{2}$ $\Rightarrow \theta =2n\pi \pm {\cos }^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$
Hence a $=\cos \varphi +i\sin \varphi ,$
where $\varphi =2n\pi \pm {\cos }^{-1}\left(\frac{\sqrt{5}-1}{2}\right);n\in Z$