Substitute . You will see what must equal.
What Plato is saying is that this equation will have many different roots depending on what a is. And that you can make one root to be anything you want by choosing a correctly. In particular, to be sure that the equation has at least one real, negative, root, you can just choose z to be any real, negative, number you want.
for example, if you chose to make that root -2 instead of -1, you would have the equation and solve that for a.
However, I see a difficulty with this. It is possible that the form (a- 3i) is supposed to indicate that a is a real number. If so, that should have been said, but if so then that makes the problem harder!
Assuming that a must be real, we can separate the equation into real and imaginary parts:
Imaginary part: .
(I have used the fact that z is a real number in separating.)
Now you can solve that second, quadratic, equation for z and see what values of a allow a negative z that satifies both equations.