# 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".