# Am I right in this conclusion

1. Jul 23, 2006

### wooby

Find all complex z = (x,y) such that $$z^2 + z + 1 = 0$$

I conclude that there is no solution set because for the real component to be 0 one must be able to solve $$x^2 + x + 1 = 0$$ and such a solution does not exist in the reals.

Am I correct or did I mess up in my algebra some where resulting in the quadratic above?

Thanks

2. Jul 23, 2006

### Data

A polynomial of degree $n$ ($\ge 1$) always has exactly $n$ roots over the complex numbers.

Do you remember something about a formula that gives solutions to quadratic equations (a "quadratic formula," perhaps?)?

3. Jul 24, 2006

### wooby

Yes, however, since x is the real component, is a complex root for x a valid answer?

4. Jul 24, 2006

### shmoe

if z=x+i*y is a root then

(x+i*y)^2+(x+i*y)+1=0

the real part of the left hand side is x^2-y^2+x+1, not x^2+x+1 (I think that's what you're saying?)

5. Jul 24, 2006

### wooby

correct, however, since we are finding solutions for 0, the complex value is zero only when y = 0 thus the -y^2 drops off the real portion.

or am I incorrect in doing that?

6. Jul 24, 2006

### shmoe

No, y does not have to be zero.

(x+i*y)^2+(x+i*y)+1=(x^2-y^2+x+1)+i*(2*x*y+y)

If x+i*y is a zero, this gives two equations:

2*x*y+y=0
x^2-y^2+x+1=0

y does not have to be zero for there to be a solution. Take x=-1/2, etc.

(setting real and imaginary parts to zero to try to find a solution here is not ideal at all, you have the quadratic formula right? You also know the fundamental theorem of algebra? Maybe not, it's what Data mentioned- a poly of degree n will have n complex roots, counting multiplicities)

7. Jul 24, 2006

### Data

Yes, you are incorrect in doing that. The imaginary part of the LHS (in shmoe's scheme) is $2xy + y$, so it depends on x too. You can solve the simultaneous equations

$$\{x^2-y^2+x+1 = 0, 2xy + y =0\}$$

and they should give you the same solution set as the quadratic formula (for $z=x+iy$).

Edit: Since this seems like it might be the source of your confusion, I will note that you should definitely only accept REAL solutions for $x$ and $y$ here (it doesn't make any difference for $z$, you will still have only at most two solutions for that. But you assume that $x$ and $y$ are real here). You will still find two solutions for $z$ (or one solution representing a root with multiplicity 2).

Last edited: Jul 24, 2006
8. Jul 24, 2006

### HallsofIvy

"Finding solutions for 0"? y is the imaginary part of the root, z, and that is definitely NOT 0. It is both real and imaginary parts of the entire
(x^2-y^2+x+1) + i(2xy+ y) that must be 0. That is, you must have
x^2- y^2+ x+ 1= 0 and 2xy+ y= 0. Yes, the latter can be factored as
y(2x+ 1)= 0 so either y= 0, in which case we must have x^2+ x+ 1= 0 which has no real roots, or2x+1= 0 so x= -1/2, in which case we must have 1/4- y^2- 1/2+ 1= 3/4- y^2= 0 so $y= \pm\frac{\sqrt{3}}{2}$: the roots are itex]z= -\frac{1}{2}\pm\frac{\sqrt{3}}{2}[/itex].

Of course, as Data originally pointed out, you could just use the quadratic formula:
$$z= \frac{-1\pm\sqrt{1- 4}}{2}= \frac{-1\pm\sqrt{2}}{2}$$

9. Jul 24, 2006

### wooby

Thanks everyone.

I can't believe I missed such an obvious solution to the Im z portion.

HallsofIvy, should you have a -3 under the radical in your last line?