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Am I right in this conclusion

  1. Jul 23, 2006 #1
    Find all complex z = (x,y) such that [tex]z^2 + z + 1 = 0[/tex]

    I conclude that there is no solution set because for the real component to be 0 one must be able to solve [tex] x^2 + x + 1 = 0 [/tex] 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. jcsd
  3. Jul 23, 2006 #2
    A polynomial of degree [itex]n[/itex] ([itex]\ge 1[/itex]) always has exactly [itex]n[/itex] roots over the complex numbers.

    Do you remember something about a formula that gives solutions to quadratic equations (a "quadratic formula," perhaps?)? :biggrin:
     
  4. Jul 24, 2006 #3
    Yes, however, since x is the real component, is a complex root for x a valid answer?
     
  5. Jul 24, 2006 #4

    shmoe

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    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?)
     
  6. Jul 24, 2006 #5
    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?
     
  7. Jul 24, 2006 #6

    shmoe

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    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)
     
  8. Jul 24, 2006 #7
    Yes, you are incorrect in doing that. The imaginary part of the LHS (in shmoe's scheme) is [itex]2xy + y[/itex], so it depends on x too. You can solve the simultaneous equations

    [tex]\{x^2-y^2+x+1 = 0, 2xy + y =0\}[/tex]

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

    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 [itex]x[/itex] and [itex]y[/itex] here (it doesn't make any difference for [itex]z[/itex], you will still have only at most two solutions for that. But you assume that [itex]x[/itex] and [itex]y[/itex] are real here). You will still find two solutions for [itex]z[/itex] (or one solution representing a root with multiplicity 2).
     
    Last edited: Jul 24, 2006
  9. Jul 24, 2006 #8

    HallsofIvy

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    "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 [itex]y= \pm\frac{\sqrt{3}}{2}[/itex]: 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:
    [tex]z= \frac{-1\pm\sqrt{1- 4}}{2}= \frac{-1\pm\sqrt{2}}{2}[/tex]
     
  10. Jul 24, 2006 #9
    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?
     
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