let a be a complex number such that |a| = 1 If the equation az² + z + 1 = 0 has a pure imaginary root, then tan(arg a) =

Let $\mathrm{z}=\mathrm{ki}\Rightarrow \overline{\mathrm{z}}=-\mathrm{ki}=-\mathrm{z}$
${\mathrm{az}}^2+\mathrm{z}+1=0\dots \left(1\right)\Rightarrow \overline{\mathrm{a}}{\overline{\mathrm{z}}}^2-\overline{\mathrm{z}}+1=0\dots \left(2\right)$
solving (1) and (2) $\mathrm{z}=\frac{-2}{\mathrm{a}-\overline{\mathrm{a}}},{\mathrm{z}}^2=\frac{-2}{\mathrm{a}+\overline{\mathrm{a}}}$
$\Rightarrow 2(\mathrm{a}+\overline{\mathrm{a}})+(\mathrm{a}-\overline{\mathrm{a}}{)}^2=0$
put $$\begin{gathered} \mathrm{a}=\mathrm{cis}\mathrm{\alpha } \mathrm{we}\mathrm{get} 4\cos \mathrm{\alpha }-4{\sin }^2\mathrm{\alpha }=0 \\ \Rightarrow {\cos }^2\alpha +\cos \alpha -1=0 \end{gathered}$$
$\cos \mathrm{\alpha }=\frac{\sqrt{5}-1}{2}\Rightarrow \tan \mathrm{\alpha }=\frac{\sqrt{2\left(\sqrt{5}-1\right)}}{\sqrt{5}-1}=\sqrt{\frac{\sqrt{5}+1}{2}}$