How to Solve the Inequality x^2 + 2ix + 3 < 0?

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The discussion centers on solving the inequality x^2 + 2ix + 3 < 0, where i represents the imaginary unit (i^2 = -1). Participants clarify that the inequality does not hold meaning in the context of complex numbers, as the ordering of complex numbers is not well-defined. However, they suggest transforming the inequality into an equality, leading to the equation x^2 + 2ix + 3 = a, where a is a real negative number. The solution for x is given by the formula x = i(-1 ± √(4 - a)), applicable when x is expressed as a pure imaginary number.

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Dacu
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Hello,
Solve inequality x^2+2ix+3&lt;0 where i^2=-1
 
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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.
 
There are some ##x## for which it does make sense though. So start by figuring out for which ##x## you obtain real numbers.
 
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Any inequality can be transformed into equality and so we can write x^2+2ix+3=a where i^2=-1 and a\in \mathbb R^-.Solving the equation is very simple ...
 
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OK... So what is the solution then?
 
Dacu said:
Any inequality can be transformed into equality and so we can write x^2+2ix+3=a where i^2=-1 and a\in \mathbb R^-.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"?
 
If x is a pure imaginary (x=iy) then the question makes sense -y^2-2y+3&lt;0\ or \ y^2+2y&gt;3
 
Hello,
Inequality solutions are given by formula:
x=i(-1\mp \sqrt{4-a}) where a\in \mathbb R ^-
 
OK... Is there any reason in particular that you created this thread?
 
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Reason:
Applications on "The fundamental theorem of algebra".
 

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