# Complex Number question

1. May 19, 2013

### mia5

1. The problem statement, all variables and given/known data

The locus of z satisfying the inequality (z + 2 i) / (2 z + i) < 1 where z = (x + i y)

2. Relevant equations

none

3. The attempt at a solution
After putting the value of z = (x + i y) , I tried to rationalize it but now I am stuck. can somebody explain how to solve it. The correct answer is x2 + y2 < 1

Thank you in advance.

Last edited: May 19, 2013
2. May 19, 2013

### SammyS

Staff Emeritus
You can't use inequalities directly with complex numbers.

Also, you need to use parentheses in your expressions to make them represent what you actually mean.

I assume you mean to be working with ( z + 2 i )/( 2 z + i)

z + 2 i / 2 z + i literally means $\ \displaystyle z+\left(2\left(\frac{i}{2}\right)z\right)+i\ .$

3. May 19, 2013

### haruspex

Do you mean |(z + 2 i) /( 2 z + i)| < 1?

4. May 19, 2013

### mia5

yes that is what I meant

5. May 20, 2013

### haruspex

Ok. Given an expression like 1/(x+iy), can you see how to get it into the form u+iv?

6. May 20, 2013

### mia5

Are you meaning to say that I have to rationalize it ?

7. May 20, 2013

### haruspex

Yes.

8. May 20, 2013

### mia5

read before you post

Well, if you had only carefully read my first post. I did rationalize it but I am stuck at that.

9. May 20, 2013

### Dick

You don't need to rationalize it. Use |a/b|=|a|/|b|. So your inequality becomes |z+2i|<|2z+i|, right? Just express both sides in terms of x and y.

10. May 20, 2013

### mia5

Why didn't I think of it before ??? But anyways thanks for the hint

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