A An inequality

1. Apr 6, 2016

Dacu

Hello,
Solve inequality $$x^2+2ix+3<0$$ where $$i^2=-1$$

2. Apr 6, 2016

ShayanJ

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. Apr 6, 2016

micromass

Staff Emeritus
There are some $x$ for which it does make sense though. So start by figuring out for which $x$ you obtain real numbers.

4. Apr 6, 2016

Dacu

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 ...

5. Apr 6, 2016

micromass

Staff Emeritus
OK... So what is the solution then?

6. Apr 6, 2016

HallsofIvy

Staff Emeritus
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. Apr 6, 2016

mathman

If x is a pure imaginary (x=iy) then the question makes sense $-y^2-2y+3<0\ or \ y^2+2y>3$

8. Apr 7, 2016

Dacu

Hello,
Inequality solutions are given by formula:
$$x=i(-1\mp \sqrt{4-a})$$ where $$a\in \mathbb R ^-$$

9. Apr 7, 2016

micromass

Staff Emeritus
OK... Is there any reason in particular that you created this thread?

10. Apr 7, 2016

Dacu

Reason:
Applications on "The fundamental theorem of algebra".