Complex Analysis prerequisite material review

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Homework Help Overview

The problem involves identifying the set of complex numbers \( z \) that satisfy the inequality \( 1 < |2z - 6| < 2 \). This falls under the subject area of complex analysis, specifically dealing with distances in the complex plane and geometric interpretations of inequalities.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the transformation of the inequality into a geometric context, discussing the implications of the distances from a point in the complex plane. There are attempts to express the solution in terms of an annulus and questions about whether to describe the set algebraically or geometrically.

Discussion Status

Participants are actively engaging with the problem, offering hints and leading questions to guide understanding. There is recognition of the need for clarity in expressing the solution, with some participants suggesting that the algebraic form may be the expected answer.

Contextual Notes

Some participants mention constraints on providing direct solutions, emphasizing the importance of reasoning through the problem rather than simply stating answers. There is also a note about the need to keep certain values away from specific points 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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