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Need proof

  1. Oct 2, 2005 #1
    How to proove that

    (x - a)(x - b)(x - c)...

    If a, b, c... are complex numbers, and none is conjugent to another the result will always be complex? Complex as is not real for those who like to complicate things...
     
    Last edited: Oct 2, 2005
  2. jcsd
  3. Oct 2, 2005 #2
    well, to start off, the first thing you need to do is to start it on your own.
     
  4. Oct 2, 2005 #3
    Well, if the equation's coefficients are not complex, but x+iY is a root, then x-iY is also a root, since we collect imaginary parts separate from real.
     
  5. Oct 2, 2005 #4
    well what do you get when you multiple one C:P(x) to a R:P(x)
     
  6. Oct 3, 2005 #5
    Well this the only thing i could proove; with two factors;

    (x - (a + bi))(x - (c + di))
    x^2 - x(a + bi + c+di) + (c+di)(a+bi)
    x?2 -x(a+c + i(b+d)) + (ca -bd + i(da + cb))

    In order for the result to be real;

    b = -d
    -da = cb

    ba=cb

    a=c

    so it is actually

    (x - (a + bi))(x - (a -bi)), otherwise the result is not real. But how to proove that it is also applicable for any amount of factors?
     
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