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Complex equation

  1. Apr 6, 2004 #1
    [tex] z^2 - (7+i)z + 24 + 7i = 0 [/tex]
    [tex] z= \frac{\ 7+i \pm \sqrt{(7+i)^2 - 4(24+7i)} }{ 2 }[/tex]
    [tex] z= \frac{\ 7+i \pm \sqrt{-47 - 14i }}{ 2 } [/tex]
    i stuck here
     
  2. jcsd
  3. Apr 6, 2004 #2

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    I don't know whether you can simplify it further than that.

    When I check the real part of what you have inside the root, I get 7^2 - 1 - 4*24, which is -48, not -47.
     
    Last edited: Apr 6, 2004
  4. Apr 6, 2004 #3
    yes, it's -48
    the answer in book is 3+4i , 4-3i
     
  5. Apr 6, 2004 #4

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    I just substituted one of the book solutions into the original equation to check whether it is valid:

    (3 + 4i) (3 + 4i) - (7 + i) (3 + 4i) + 24 + 7i = 0 + 0i.

    So that book solution seems to be correct.

    I also checked the other book solution, and it is also correct.

    I do not know how you get from your form to the book form.
     
    Last edited: Apr 6, 2004
  6. Apr 6, 2004 #5

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    To simplify sqrt(-48-14i), solve for real numbers a and b such that:

    (a + bi)(a+bi)= -48-14i,

    so that a^2 - b^2 = -48 and 2ab = -14.

    You may be able to figure it out by doing that.

    The next step could be that since a=-7/b,

    (-7/b)^2 - b^2 = -48.
     
  7. Apr 6, 2004 #6
    get it!!
    Thank~~~
     
  8. Apr 6, 2004 #7

    HallsofIvy

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    A standard way of handling [itex]\sqrt{-48-14i}[/itex] is to convert to polar form and use DeMoivre's formula: this happens to have r= 50 and [theta]= 0.284 radians.
    The square root will have r= [itex]\sqrt{50}= 5\sqrt{2}[/itex] and theta= 0.142 radians or 3.283 radians. That is, the square roots are [itex]5\sqrt{2}(cos(0.142)+ i sin(0.142))[/itex]= 7+ i and [itex]5\sqrt{2}(cos(3.283)+ i sin(3.283))[/itex]= -7- i.
     
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