Locus of z satisfying (z + 2i)/(2z + i) < 1

[mia5]

Homework Statement

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

Homework Equations

none

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.

 

[SammyS]

Homework Statement

The locus of z satisfying the inequality z + 2 i / 2 z + i is less than 1 where z = x + i y

Homework Equations

none

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 is less than 1

Thank you in advance.

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

 

[haruspex]

z + 2 i / 2 z + i is less than 1

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

 

[mia5]

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

yes that is what I meant

 

[haruspex]

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

 

[mia5]

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

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

 

[haruspex]

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

Yes.

 

[mia5]

read before you post

Yes.

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

 

[Dick]

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

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.

 

[mia5]

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.

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