Complex Plane

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  • #1
danago
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Given that [tex]|z-1+i| \le 1[/tex], find the maximum and minimum value of |z| and Arg(z).

I realise that the equation given defines the interior of a circle of radius 1 centered at (1,-1), which includes the circumference.

For the first part of the question, i am able to represent the equation graphically. From what i understand, |z| is the distance from the origin to any point lying on or within the circle. If this is the case, i can see the minimum and maximum points, but im not too sure on how to calculate their locations.

For the next part, finding the extreme values of Arg(z), i just read straight from my graph and said that the minimum is [tex]-\pi/2[/tex] and the maximum is 0. Is that right?

Thanks in advance,
Dan.
 

Answers and Replies

  • #2
mjsd
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from description, it appears that this is your task:
given the set of points defined by [tex]\{z:|z-1+i|\leq 1\}[/tex] find the complex numbers z such that the vector going from origin to z has max/min length. likewise for angle (so -p1/2 and 0 seem wrong). but then again you said you read straight from your graph, how does your graph look like, or how you derived it? (to help with pin-pointing potential mistakes)


EDIT: sorry my mistake
 
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  • #3
danago
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I simply drew a circle centered at (1,-1) with radius 1. Both the x and y axis are tangental to the circle. That pretty much explains what ive drawn.

My book says the argument of a complex number should be defined within -pi to pi, which is how i got 0 and -pi/2, since the circle touches the positive x axis and the negative y axis.
 
  • #4
Dick
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Ok, seems like your picture is fine. For |z|, how far is the center of the circle from the origin? Now how far are the closest and farthest points from the center of the circle?
 
  • #5
danago
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Ah ok thats a good way to think about it. The distance from the origin to (1,-1) is [tex]\sqrt{2}[/tex], plus another 1 unit (the circles radius) gives a maximum distance of [tex]\sqrt{2}+1[/tex]. The minimum distance will just then be [tex]\sqrt{2}-1[/tex]. Am i right?
 
  • #6
Dick
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Absolutely.
 
  • #7
danago
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Alright thanks for the help :smile: What about the argument of z? Was i right with that?
 
  • #8
HallsofIvy
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Yes, you were.
 
  • #9
danago
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Alright, thanks alot :smile:
 

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