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Sum of two square and complex quadratic

  1. Dec 5, 2009 #1
    I actually have two questions, both are in the complex plane.

    1. The problem statement, all variables and given/known data

    Q1: Express 962 as a sum of two squares (Hint: 962 = (13)(74)
    Q2. Given z,a belong to C (complex). Find a such that the roots of the equation [tex] z^2 + az + 1 = 0 [/tex] have equal absolute values (or modulus).



    2. Relevant equations
    Well this is about the complex plane so for the first one I think most probably its Euler's Identity. As for the second it could be anything.


    3. The attempt at a solution

    Q1: It is a sum of squares so the best I can think of is that 962 is a modulus of a vector i.e. z belongs to C, |z| = 962. Because z = a + ib so [tex]962^2 = a^2 + b^2 [/tex]. Then I try writing z in polar so we have [tex]z = |z|e^i^x [/tex] , where x is the argument of z. But x = arctan b/a . So I somewhat got stuck in a loop.

    Q2: We can't use the auxiliary equation because a is also complex. So it doesn't work. I tried expanding both z and a only to get stuck eventually in an overly complicated equation. Any suggestions on just how to start it?
     
  2. jcsd
  3. Dec 5, 2009 #2

    HallsofIvy

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    The problem asks you to find a and b such that [itex]a^2+ b^2= 962[/itex], NOT [itex]962^2[/itex]! Although it doesn't have anything to do with this hint, one way to do it is by "brute strength": Is 962- 1 a square? 962- 4? 962- 9?...

    I have no idea what you mean by this. What "auxiliary equation? And the real numbers are a subset of the complex numbers- saying that "a is a complex number" doesn't mean it can't be real. There is an obvious real number solution for a.

     
  4. Dec 5, 2009 #3
    The auxiliary equation is [tex] x = \frac{-b \frac{+}{-} \sqrt{4ac - b^2}}{2a}[/tex] where [tex]ax^2+bx+c=0[/tex]. That's what I know it as anyways. Anyways, even the teacher said you can't do it that way. But if you can find a way, then I'd be more than happy.

    You're right, I'm not saying a can't be real, but we also cannot make the assumption that it is purely real or imaginary. Why is there an obvious real number solution?

    As for the sum of squares: I tried by brute strength, which did work, but its time consuming and has nothing to do with the hint or what we're studying. It turned out to be [tex] 962 = 11^2+29^2[/tex].
     
    Last edited: Dec 5, 2009
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