Complex Analysis prerequisite material review

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SUMMARY

The discussion revolves around solving the inequality set defined by the complex numbers satisfying the condition ##1<\vert 2z-6\vert <2##, where ##z\in\Bbb{C}##. The solution involves transforming the inequality into a geometric interpretation, identifying the annulus centered at (3,0) with inner radius 1/2 and outer radius 1. The participants conclude that the algebraic representation of the solution can be expressed as ##H\cap O=\{z=x+iy\in\Bbb{C}:\frac{1}{4}<(x-3)^2+y^2<1\}##, which accurately describes the set of points in the complex plane.

PREREQUISITES
  • Understanding of complex numbers and their geometric representation
  • Familiarity with inequalities in the context of the complex plane
  • Knowledge of algebraic manipulation involving complex variables
  • Basic concepts of annuli and circles in two-dimensional geometry
NEXT STEPS
  • Study the geometric interpretation of complex inequalities
  • Learn about annuli and their properties in the complex plane
  • Explore algebraic representations of geometric shapes in complex analysis
  • Investigate the implications of distance in the complex plane for solving inequalities
USEFUL FOR

Students of complex analysis, mathematicians focusing on geometric interpretations, and educators seeking to enhance their understanding of inequalities in the complex plane.

Terrell
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Homework Statement


Identify the set of points satisfying ##1<\vert 2z-6\vert <2## such that ##z\in\Bbb{C}##.
My pre-caculus is very rusty, so I am not sure if I am doing this correctly.

Homework Equations


##x^2 +y^2= r^2##
##\forall z,z'\in\Bbb{C}, \vert zz'\vert =\vert z\vert\vert z'\vert##

The Attempt at a Solution


Let ##z=x+iy##. Then ##\vert 2z-6\vert =\vert 2(z-3)\vert=\vert 2\vert\vert z-3\vert## and \begin{align}1<\vert 2z-6\vert<2\Longleftrightarrow \frac{1}{2}<\vert z-3\vert <1\end{align}
We want complex numbers ##z## that are at least ##1/2## units away from ##3## and at most ##1## unit away from ##3##. Let ##H=\{z\in\Bbb{C}:\frac{1}{2}<\vert z-3\vert\}## and ##O=\{z\in\Bbb{C}:\vert z-3\vert <1\}##. We want to find the conditions of all ##z\in\Bbb{C}## such that ##z\in H\cap O##. Since ##\Bbb{R}^2\cong\Bbb{C}##, then the general form of elements in ##H## can be found by solving: ##(x-3)^2 +y^2>(\frac{1}{2})^2\Longrightarrow y>\pm\sqrt{\frac{1}{4}-(x-3)^2}## and for ##O## we solve: ##(x-3)^2+y^2<1^2\Longrightarrow y<\pm\sqrt{1-(x-3)^2}##. Therefore, we have \begin{align}\pm\sqrt{\frac{1}{4}-(x-3)^2}<y< \pm\sqrt{1-(x-3)^2}\Longleftrightarrow 1/4<y<1\end{align}
Hence, ##H\cap O=\{z=x+iy\in\Bbb{C}:\pm\sqrt{\frac{1}{4}-(x-3)^2}<y< \pm\sqrt{1-(x-3)^2}\quad \land\quad 1/4<y<1 \}##
 
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You can easily describe the set as an annulus centered at TBD with inner radius TBD and outer radius TBD.
Fill in the blanks and do a sanity check on your answer. You are keeping y away from 0. Does that seem right?

PS. Do you think that the expected answer is in terms of (x,y), or would they expect you just to describe it geometrically in terms of an annulus?
 
FactChecker said:
TBD
What is TBD?
 
Terrell said:
What is TBD?
On homework problems, I am not allowed to do more than ask leading questions and give hints to redirect you. You should be able to fill in the 'TBD' (To Be Determined) blanks. If not, you may want to think about the problem more in terms of geometric position and distances in the complex plane.
 
FactChecker said:
to think about the problem more in terms of geometric position and distances in the complex plane
I think it is all the complex numbers between the circle of radius 1/2 units and circle of radius 1 centered at coordinate (3,0) right?
 
Terrell said:
I think it is all the complex numbers between the circle of radius 1/2 units and circle of radius 1 centered at coordinate (3,0) right?
Right. That should make you doubt your answer above, which keeps y away from the x-axis (y=0).

PS. Your original equation 1 is the best algebraic description of the set. I think that may be what they expected as an answer. Or the statement you gave just now in terms of the circles (annulus).
 
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FactChecker said:
I think that may be what they expected as an answer.
I am starting to think so too. I'm having difficulty finding a single expression to describe it and I'm not sure if I have forgotten how or there is none.
 
can I just simply state that ##H\cap O=\{z=x+iy\in\Bbb{C}:\frac{1}{4}<(x-3)^2+y^2<1\}##? Thanks!
 
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Terrell said:
can I just simply state that ##H\cap O=\{z=x+iy\in\Bbb{C}:\frac{1}{4}<(x-3)^2+y^2<1\}##? Thanks!
Good answer.
 
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