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A An inequality

  1. Apr 6, 2016 #1
    Solve inequality [tex]x^2+2ix+3<0[/tex] where [tex]i^2=-1[/tex]
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  3. Apr 6, 2016 #2


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    The ordering you mean, that is well-defined for real numbers, is not defined for complex numbers. So your inequality doesn't make sense.
  4. Apr 6, 2016 #3
    There are some ##x## for which it does make sense though. So start by figuring out for which ##x## you obtain real numbers.
  5. Apr 6, 2016 #4
    Any inequality can be transformed into equality and so we can write [tex]x^2+2ix+3=a[/tex] where [tex]i^2=-1[/tex] and [tex]a\in \mathbb R^-[/tex].Solving the equation is very simple ...
  6. Apr 6, 2016 #5
    OK... So what is the solution then?
  7. Apr 6, 2016 #6


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    Yes, it is. Do you understand that the inequality you originally post makes no sense?

    You say "any inequality can transformed into an equality". Of course, you can just replace "<" or ">" with "=" but that is not what I would call "transforming"?
  8. Apr 6, 2016 #7


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    If x is a pure imaginary (x=iy) then the question makes sense [itex]-y^2-2y+3<0\ or \ y^2+2y>3[/itex]
  9. Apr 7, 2016 #8
    Inequality solutions are given by formula:
    [tex]x=i(-1\mp \sqrt{4-a})[/tex] where [tex]a\in \mathbb R ^-[/tex]
  10. Apr 7, 2016 #9
    OK... Is there any reason in particular that you created this thread?
  11. Apr 7, 2016 #10
    Applications on "The fundamental theorem of algebra".
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