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?
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.
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.
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.
An ideea: The inequation x2+2ix+3<0 is equivalent to the equation x2+2ix+3=a where i2=−1 and a∈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 ...
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?
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.
To rephrase the question such that it actually makes sense... (I use z here because, well, convention)
Given z∈C such that z2+2iz+3 is real, find all z∈C such that Re(z2+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.