How Do You Separate Complex Equations into Real and Imaginary Parts?

[k3N70n]

Homework Statement

Write z^3 + 5 z^2 = z + 3i as two real equations

Homework Equations

z=a+bi?

The Attempt at a Solution

I've been just playing around with this. I expanded, grouped the real and imaginary parts. I'm really just think I'm groping around desperately in the dark.
I think I'm just missing something basic. Any hints would be greatly appreciated. Until then I'll be having a head butting contest with the sidewalk.

 

[rock.freak667]

I would just expand out and group as well...as in the form z=A+Bi both A and B are real so...

 

[k3N70n]

Okay.
So far I did this, I don't think there's any basic algebraic errors though it wouldn't be the first time.
(a+bi)^3 + 5(a+bi)^2 = a + bi + 3i
==> (a^3 +3 a^2 bi - 3ab^2-b^3 i) + (5a^2 +10abi-db^2) = a+bi+3i
==> (a^3-3ab^2+5a^2-5b^2-a)+(3a^2b-b^3+10ab-b-3)i=0
so somehow I have to use some mathematical magic to make that i dissappear? Is this just a question that I need to play around with algebraically for a while?

Alright Kenton vs the Sidewalk round two

 

[Dick]

Assuming you did that right, then x+iy=0 (for x and y real) only if x=0 and y=0. That's two real equations.

 

[rock.freak667]

And also you did not really need to bring everything over to one side to make it equal to zero. Remember that two complex numbers are equal iff their real and imaginary parts are the same.

 

[HallsofIvy]

Okay.
So far I did this, I don't think there's any basic algebraic errors though it wouldn't be the first time.
(a+bi)^3 + 5(a+bi)^2 = a + bi + 3i
==> (a^3 +3 a^2 bi - 3ab^2-b^3 i) + (5a^2 +10abi-db^2) = a+bi+3i
==> (a^3-3ab^2+5a^2-5b^2-a)+(3a^2b-b^3+10ab-b-3)i=0
so somehow I have to use some mathematical magic to make that i dissappear? Is this just a question that I need to play around with algebraically for a while?

Alright Kenton vs the Sidewalk round two

It's a problem where you need to know the definitions. Under what conditions on x and y is x+ iy= 0?