Question: If $a$ is a complex number such that $\vert a\vert=1$, then find the values of $a$ such that the equation $az^2+z+1=0$ has one purely imaginary root.
The equation can be written as $$z^2+\bar{a}z+\bar{a}=0$$ $$z=-\frac{\bar{a}}{2}\pm\sqrt{\frac{\bar{a}^2}{4}-\bar{a}}$$ that did not work. So I tried using the sum andproduct of the roots:
Let the roots be $x$ and $ki$.
$$x+ki=-\bar{a}$$ $$xki=\bar a$$ From the second equation, $$x=-\frac{\bar a}{k}i$$
Substituting $x$ in first equation, $$k^2-k\bar ai-\bar a=0$$ The above equation should have a real solution for $k$ On using the discriminant formula, I am not getting the correct answer.