Complex Number: What's the set?

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SUMMARY

The discussion centers on the set defined by the inequality { z ∈ ℂ | |z|² ≥ z + z̅ }. Participants derive that this leads to the condition a² + b² ≥ 2a, which can be rewritten as (a - 1)² + b² ≥ 1, representing a circle of radius 1 centered at (1, 0). The geometric interpretation reveals that the set consists of points outside or on this circle, indicating that the length of the complex number is greater than or equal to twice its real part. This conclusion provides a clear visualization of the set in the complex plane.

PREREQUISITES
  • Understanding of complex numbers and their representation in the form z = a + ib
  • Familiarity with inequalities and geometric interpretations in the Cartesian plane
  • Knowledge of completing the square in algebra
  • Basic concepts of circles in coordinate geometry
NEXT STEPS
  • Explore the geometric properties of complex numbers in the Argand plane
  • Learn about inequalities involving complex numbers and their graphical representations
  • Study the implications of transformations in the complex plane, particularly translations and scaling
  • Investigate advanced topics such as complex analysis and its applications in geometry
USEFUL FOR

Mathematicians, students studying complex analysis, educators teaching geometry, and anyone interested in the geometric interpretation of inequalities involving complex numbers.

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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