Complex Number: What's the set?

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Homework Help Overview

The discussion revolves around the set of complex numbers defined by the inequality \{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}. Participants are exploring the implications of this inequality, particularly in terms of geometric interpretation and visualization.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to manipulate the inequality by expressing the complex number in terms of its real and imaginary components, leading to the expression a^2 + b^2 ≥ 2a. They explore the geometric interpretation of this inequality and question how to visualize the resulting set of points.

Discussion Status

There is ongoing exploration of the geometric representation of the inequality, with some participants suggesting it describes a circle and others confirming this interpretation. Multiple perspectives on visualization are being discussed, but no consensus has been reached on the most effective description.

Contextual Notes

Participants are considering the implications of the inequality and its geometric representation, including the constraints of the problem and the nature of the set defined by the inequality.

latentcorpse
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What's the set \{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}?

I've set z=a+ib and found a^2 + b^2 \geq 2a \Rightarrow b^2 \geq a(2-a)

I'm not sure how to interpret this geometrically ie what it looks like?

I suppose it is the set of vectors whose length is bigger than twice their real part. I guess if I take the square root then I find \{ b \geq \pm \sqrt{a} \} \cap \{ b \geq \pm \sqrt{2-a} \}

How do we draw this?

Thanks.
 
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latentcorpse said:
What's the set \{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}?

I've set z=a+ib and found a^2 + b^2 \geq 2a \Rightarrow b^2 \geq a(2-a)
##a^2 - 2a + b^2 \geq 0##
Complete the square in the a terms, and see what you get.
latentcorpse said:
I'm not sure how to interpret this geometrically ie what it looks like?

I suppose it is the set of vectors whose length is bigger than twice their real part. I guess if I take the square root then I find \{ b \geq \pm \sqrt{a} \} \cap \{ b \geq \pm \sqrt{2-a} \}

How do we draw this?

Thanks.
 
latentcorpse said:
What's the set \{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}?

I've set z=a+ib and found a^2 + b^2 \geq 2a \Rightarrow b^2 \geq a(2-a)

I'm not sure how to interpret this geometrically ie what it looks like?

I suppose it is the set of vectors whose length is bigger than twice their real part. I guess if I take the square root then I find \{ b \geq \pm \sqrt{a} \} \cap \{ b \geq \pm \sqrt{2-a} \}

How do we draw this?

Thanks.

Interpret ##a^2 -2a + b^2 \geq 0##.
 
so
(a-1)^2+(b-0)^2-1 \geq 0

(a-1)^2+(b-0)^2 \geq 1

so circle radius 1 centre (1,0)?
 
Mark44 said:
Yes.

what was wrong with saying the length was bigger than twice the real part?
 
Nothing, but which description is easier to visualize?
 
latentcorpse said:
so
(a-1)^2+(b-0)^2-1 \geq 0

(a-1)^2+(b-0)^2 \geq 1

so circle radius 1 centre (1,0)?
(a- 1)^2+ b^2= 1
is a circle of radius 1 with center at (1, 0).

(a- 1)^2+ b^2\ge 1
is the set of points outside that circle.
 
HallsofIvy said:
(a- 1)^2+ b^2= 1
is a circle of radius 1 with center at (1, 0).

(a- 1)^2+ b^2\ge 1
is the set of points outside that circle.
The inequality represents the set of points on the circle or outside it.
 

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