Problem
Define constant $a$ in way that $x^2+(3a+1)x+81=0$ solutions are complex.
After that define $a$ in way that the solutions are strictly imaginary (when real part is $0$)
Attempt to solve
In a way real solutions for this are also complex. Real numbers are subset of complex numbers hence all real solutions are also complex. I think this was intended to interpret as solutions that are form $z=a+ib$ when $(a,b) \in \mathbb{R}, z \in \mathbb{C}$, $b \neq 0$ so solution is complex when it has imaginary component.
$$ x^2+(3a+1)x+81=0 $$
By utilizing quadratic formula we can use discriminant to define for what constant $a$ value equation has only complex solutions.
$$ D=(3a+1)^2-4\cdot 81 $$
if $D<0$ all solutions have to have imaginary part hence they are complex.
Our discriminant is form of a parabola if we would use $a$ as variable.
$$ (3a+1)^2=9a^2+6a+1 $$
if we compute when $D=0$ we can figure out when $D < 0$
$$ 9a^2+6a-323 = 0 $$
$$ a = \frac{ -6 \pm \sqrt{6^2-4 \cdot (-323) \cdot 9} }{ 2 \cdot 9 } $$
$$ a = \frac{ -6 \pm 108 } { 18 } $$
$$ a_1 = \frac{ -19 }{ 3 }, a_2 = \frac{ 17 }{ 3 } $$
So we know that $D < 0$ when:
$$ \frac{-19}{3}<a<\frac{17}{3} $$
Now only problem is i don't know if my solution is valid and how do you define a when solutions have to be strictly imaginary ?
Best Answer
Second part
$$x^2+(3a+1)x+81=0\implies x=\frac{-(3a+1)\pm \sqrt{(3a+1)^2-4\cdot 81}}{2}.$$
The roots are strictly imaginary if and only if $3a+1=0.$
Edit
The roots are
$$\dfrac{-(3a+1)\pm \sqrt D}{2}$$ where $D$ is a real number. So
$$\sqrt{D}=\pm i\sqrt{|D|}$$ if $D<0.$ Thus
$$\Re \{\dfrac{-(3a+1)\pm \sqrt D}{2}\}=\dfrac{-(3a+1)}{2}. $$
If $D\ge 0$ the solutions are real.