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Complex Quadratic Formula

  1. Feb 7, 2008 #1
    [SOLVED] Complex Quadratic Formula

    The problem statement, all variables and given/known data
    Let a, b and c be complex numbers with a not equal to 0. Show that the solution of [itex]az^2 + bz + c = 0[/itex] are [itex]z_1, z_2 = (-b \pm \sqrt(b^2 - 4ac))/(2a)[/itex]

    The attempt at a solution
    I'm assuming z is also complex. Multiplying the equation by 4ac and completing the square yields [itex](2az + b)^2 = b^2 - 4ac[/itex] There are two complex numbers, [itex]w_1, w_2[/itex], that are the square roots of [itex]b^2 - 4ac[/itex] so setting [itex]2az_j + b = w_j[/itex] and solving for [itex]z_j[/itex] yields

    [tex]z_j = \frac{-b + w_j}{2a}[/tex]

    for j = 1, 2. The section where I got this problem from does not define the square root of a complex numbers; it only talks about the roots of complex numbers. Would it be correct to stop here and consider the problem solved?
     
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  3. Feb 7, 2008 #2

    HallsofIvy

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    How about just putting
    [tex]\frac{-b+ \sqrt{b^2- 4ac}}{2a}[/tex]
    and
    [tex]\frac{-b- \sqrt{b^2- 4ac}}{2a}[/tex]
    into the equation az2+ bz+ c= 0 and see what happens?

    (I am puzzled about " does not define the square root of a complex numbers; it only talks about the roots of complex numbers". How does it talk about the roots of complex numbers without defining them? In any case, I would think the definition is obvious: a complex number, z, is a square root of the complex number a if and only if z2= a.)
     
  4. Feb 7, 2008 #3
    My confusion stems from how the radical symbol extends to complex numbers. So the notation [itex]\sqrt{z}[/itex] is the principal square root of z? Which of the two roots is the principal though? The one with the same signs in the real and imaginary part as z?
     
  5. Feb 7, 2008 #4

    rock.freak667

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    [itex]\sqrt{z}[/itex]

    z is a complex number of the form a+bi


    [itex]\sqrt{z}[/itex] is different from a complex number,z, being a root of a quadratic.
     
  6. Feb 7, 2008 #5
    Yes, I know it is different. That doesn't answer my question.
     
  7. Feb 7, 2008 #6

    NateTG

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    Pick one - the positive one, or the one with a positive imaginary part.

    The roots will still be [itex]\pm \sqrt{z}[/itex].
     
  8. Feb 7, 2008 #7

    Avodyne

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    Just as for real numbers, every complex number has two square roots. If [itex]z=re^{i\theta}[/itex], with [itex]r[/itex] real and positive and [itex]-\pi\le\theta<\pi[/itex], the two square roots are [itex]+\sqrt{r}e^{i\theta/2}[/itex] and
    [itex]-\sqrt{r}e^{i\theta/2}[/itex]. The first of these is usually taken to be the principal root. There is a "branch cut" along the negative real axis; the square root of z changes sign abruptly as z moves across the negative real axis. Although this is the conventional choice, the branch cut could instead be chosen to lie along any line (straight or curved) from 0 to infinity.
     
  9. Feb 7, 2008 #8

    Dick

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    Wikipedia says put the branch cut along the negative real axis, so define arg(z) to be between -pi and pi and then define sqrt(z) to be the root with argument arg(z)/2. But that's not the only choice. I think it's better to define what you mean rather than assuming that 'everyone knows'.
     
  10. Feb 8, 2008 #9

    HallsofIvy

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    Since the complex numbers are not an "ordered" field there is no "principle" root except, as Dick says, by some arbitrary choice- that in general cannot be extended to other roots.

    And, it is not necessary for the problem you stated. Your problem involves both roots and it doesn't matter which you consider "principle". The only thing you need to know about [itex]\sqrt{b^2- 4ac}[/itex] is that [itex]\left(\sqrt{b^2- 4ac}\right)^2= b^2- 4ac[/itex].
     
  11. Feb 8, 2008 #10
    OK. I think I'm satisfied with the explanations. Thank you everyone.
     
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