Complex Analysis prerequisite material review

  • #1
317
26

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 \}##
 

Answers and Replies

  • #2
FactChecker
Science Advisor
Gold Member
6,053
2,340
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?
 
  • #4
FactChecker
Science Advisor
Gold Member
6,053
2,340
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.
 
  • #5
317
26
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?
 
  • #6
FactChecker
Science Advisor
Gold Member
6,053
2,340
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).
 
  • Like
Likes Terrell
  • #7
317
26
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.
 
  • #8
317
26
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!
 
  • Like
Likes FactChecker
  • #9
FactChecker
Science Advisor
Gold Member
6,053
2,340
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.
 
  • Like
Likes Terrell

Related Threads on Complex Analysis prerequisite material review

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
916
Replies
1
Views
4K
Replies
5
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
8
Views
905
Replies
11
Views
1K
Replies
7
Views
2K
Replies
5
Views
2K
Top