z = x + iy, which satisfies
z^2 + z^ -2 = 8
Find the locus of P in terms of x and y.
Is the general aproach to get a equantion in x and y?
If so, i tried this and ive got z^4 - 8z^2 + 1 = 0
Do i just sub in x + iy as z and expand to get locous?
If not what do i do?
No, I don't believe that that's quite right. Going back to and setting , we get .
Comparing real- and imaginary parts of the left and right hand side we see that we must have and .
But is not possible, because then the second of these two equations would not have a (real) solution y.
Setting to satisfy the first equation gives the solutions of the second equation.
Hence I get four isolated points (and not four vertical lines).
As I wrote, it is just the four points on the real axis. This agrees with the solution Soroban gave you, with the small difference that y cannot be arbitrary, but must be =0.
If you take , for example, which would lie on one of those vertical axes Soroban found, you get that the left hand side of the original equation evaluates to , which is rather too far away from the value of the right hand side, .