An inequality

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  • Thread starter Dacu
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  • #1
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Hello,
Solve inequality [tex]x^2+2ix+3<0[/tex] where [tex]i^2=-1[/tex]
 
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  • #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.
 
  • #3
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There are some ##x## for which it does make sense though. So start by figuring out for which ##x## you obtain real numbers.
 
  • #4
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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 ...
 
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  • #5
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OK... So what is the solution then?
 
  • #6
HallsofIvy
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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 ...
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"?
 
  • #7
mathman
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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]
 
  • #8
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Hello,
Inequality solutions are given by formula:
[tex]x=i(-1\mp \sqrt{4-a})[/tex] where [tex]a\in \mathbb R ^-[/tex]
 
  • #9
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OK... Is there any reason in particular that you created this thread?
 
  • #10
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Reason:
Applications on "The fundamental theorem of algebra".
 

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