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Homework Help: Complex solutions

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img238.imageshack.us/img238/1381/complexle5.jpg [Broken]


    3. The attempt at a solution

    I figure if i just find 1 zero of the equation i can use its conjugate as the other zero, then I guess it's complete.

    I know this:

    where a is an element of C (Complex)

    (z - a) (z - a(conj)) = z^2 - 2(Re a)z + |a|^2

    so.. using this

    2(Re a) = 1 + 3i
    and
    |a|^2 = -1 + 3i/4

    But what to do from here?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 3, 2008 #2
    Hmm perhaps I have to find the discriminant:

    which is -4 + 3i

    so z = (-(1 + 3i) +- i(sqrt (4 - 3i)))/2

    :/ doesn't look too correct
     
  4. Mar 3, 2008 #3

    tiny-tim

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    Homework Helper

    quadratic equation …

    This is an ordinary quadratic equation.

    Just solve it the ordinary way. :smile:

    (The only difficulty is that you'll have to calculate the square root of a complex number.)
     
  5. Mar 4, 2008 #4

    HallsofIvy

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    No, that's not true. The two roots of a quadratic equation must be conjugates only if the equation has all real coefficients.

    but that's impossible: 2(Re a) must be a real number.
    Again, the roots of this equation are not complex conjugates.

    Once more, that's impossible. |a| is a real number, its square is a real number.

    Use the quadratic formula or complete the square.
     
    Last edited by a moderator: May 3, 2017
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