A complex number question

One way to approach the problem is to solve the first equation to find all possible values for (z-a). Since (z-a) is raised to the third power in this equation, we expect to find three possible values. This will give you three equations all of the form z-a=#.

I'm not completely sure if I understand what you mean by the second equation, but I'm assuming that in your notation z1, z2, and z3 are the three possible values for z. So the three equations you got earlier are z1-a=#, z2-a=#, and z3-a=# (# stands for three different numbers). I'm also assuming that a is some complex constant. (Some particular number of the form x+y*i.)

Using the three equations you got from solving the first equation and also using z1*z2*z3=-9 you can solve for z1, z2, z3, and a. This is because you now have four equations and four unknowns:
z1-a=#
z2-a=#
z3-a=#
z1*z2*z3=-9

 

I presume that when you used d'Moivre's formula you found that
z- a= 2, or [itex]-1+ i\sqrt{3}[/itex], or [itex]-1-i\sqrt{3}[/itex].
Then z= 2+ a, or [itex]a-1+i\sqrt{3}[/itex], or [itex]a-1- i\sqrt{3}[/itex].
Multiplying those together and setting them equal to -9, you get a very simple equation for a!

 

It's not impossible to solve. The complex numbers are algebraically closed.

 

The equation you have on your file is exactly what I got: (a+2)(a²- 2a+ 4)= -9. Did you multiply it out? It is exactly a³+ 8= -9 or a³= -17. Surely you can solve that- just take the cube root.