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An inequation

To solve the inequation x2+2ix+3<0x^2+2ix+3<0 where i2=1i^2=-1.
(edited 10 months ago)
Reply 1
To solve the inequation x2+2ix+3<0x^2+2ix+3<0 where i2=1i^2=-1.

Complex numbers are not ordered, so it doesnt really make much sense. If its a textbook question, can you upload a picture of it?
Reply 2
No idea if it is correct, but chat GPT has an answer
Reply 3
Original post by hotpud
No idea if it is correct, but chat GPT has an answer

Interested in what it says, can you post it? Id guess its restricting x to be imaginary?
(edited 10 months ago)
Reply 4
Original post by mqb2766
Complex numbers are not ordered, so it doesnt really make much sense. If its a textbook question, can you upload a picture of it?
I am the author of the problem and obviously I have the solution! Although there is no order relation in the set of complex numbers with non-zero imaginary part, this inequation can be solved very easily if we make a transformation of this inequation.What can we transform any inequation into?
Reply 5
I am the author of the problem and obviously I have the solution! Although there is no order relation in the set of complex numbers with non-zero imaginary part, this inequation can be solved very easily if we make a transformation of this inequation.What can we transform any inequation into?
So are you restricting x to a certain domain or ...? If youre the author of the question, it would help to say in the OP.
(edited 10 months ago)
Reply 6
Original post by mqb2766
Interested in what it says, can you post it? Id guess its restricting x to be imaginary?

It all came back a little. You split the equation into the imaginary part and the non-imaginary part. Can't copy and paste as lots of characters didn't copy. Why not try it yourself? Use x^2 for x squared.
Reply 7
Original post by hotpud
It all came back a little. You split the equation into the imaginary part and the non-imaginary part. Can't copy and paste as lots of characters didn't copy. Why not try it yourself? Use x^2 for x squared.

Thats one interpretation of the question.
Reply 8
An ideea:
The inequation x2+2ix+3<0x^2+2ix+3<0 is equivalent to the equation x2+2ix+3=ax^2+2ix+3=a where i2=1i^2=-1 and aR,a<0a\in \mathbb R , a<0.
Reply 9
An ideea:
The inequation x2+2ix+3<0x^2+2ix+3<0 is equivalent to the equation x2+2ix+3=ax^2+2ix+3=a where i2=1i^2=-1 and aR,a<0a\in \mathbb R , a<0.

Wbf, you are restricting x to the imaginary numbers (or the imaginary part of the LHS=0) to make the question well defined as per #3. Why not just be clear about it in the OP? If so, just factorise the left of the original inequality as usual, or ...
(edited 10 months ago)
Reply 10
Original post by mqb2766
Wbf, you are restricting x to the imaginary numbers (or the imaginary part of the LHS=0) to make the question well defined as per #3. Why not just be clear about it in the OP? If so, just factorise the left of the original inequality as usual, or ...
What does "wbf" mean?!?Are you saying that my idea is not good?
Reply 11
What does "wbf" mean?!?Are you saying that my idea is not good?

Ill drop out at this point, but the OP could be interpreted as
Re(#)<0 and Im(#)<0
or
Re(#)<0 and Im(#)=0
or... It sounds like you meant the latter and it reduces to a simple quadratic, which can be solved in a few, similar ways. What youre proposing is introducing a slack variable to solve it.
(edited 10 months ago)
To rephrase the question such that it actually makes sense... (I use z here because, well, convention)

Given zCz\in\mathbb{C} such that z2+2iz+3z^2+2iz+3 is real, find all zCz\in\mathbb{C} such that Re(z2+2iz+3)<0\text{Re}(z^2+2iz+3) < 0.

It's conceptually wrong to use the inequality sign in complex land, since as mqb mentions, there is no order there (e.g. Is 1 or i bigger? That's not a meaningful question).

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As a sidenote, if you've concocted this question yourself with the solution, then... uh... is this post just for fun? That's fine with me - I'm just curious.
(edited 10 months ago)

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