Complex quadratic equation having one purely imaginary root

Question

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.

Answer 1

Your last equation can be obtained by just plugging $ki$ into the original equation. Then starting from here, let $a=a_1+a_2 i$ and equate the real part and imaginary part of the equation with $0$, respectively.

You will then get $$a_1=\frac{1}{k^2}\\ a_2=\frac{1}{k}$$

Now recall that $|a|=1$. Using this you can solve for $k$.

Answer 2

If $a(ir)^2+ir+1=0$ then $r$ cannot be $0$ and therefore

$$a=\frac{ir+1}{r^2}.$$

If $a$ is on the unit circle then the denominator and numerator must have equal norm, so

$$r^2+1=r^4$$

determines the four possible values of $r.$