anonymous
  • anonymous
Draw the set S of complex numbers such that \[|z-2i|<2,,, \Im (z)=1\] What is the argument when \[z \in S\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[|a-i|<2\]
anonymous
  • anonymous
But what does that say on a graph?
anonymous
  • anonymous
-2

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experimentX
  • experimentX
says it lines in the region inside a circle with radius 2 and center 0,2
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=|a-i|%3C2
anonymous
  • anonymous
It's no circle?
anonymous
  • anonymous
Is it because the Im(z) is constant at 1 and -2< Re(z) < 2, so it just varies in the Re(z) plane?
experimentX
  • experimentX
so you want z such that it's imaginary part is constant?
anonymous
  • anonymous
No i think I'm a bit wrong in my thoughts
experimentX
  • experimentX
put z = x + i in that equation |x + i - 2i| < 2 x^2 + 1^2 < 4 x^2 < 3 |x| < sqrt(3) -3 < x < 3
anonymous
  • anonymous
Is it not Re betweem -2 and 2 ?
experimentX
  • experimentX
sorry ... sqrt(3) .. lol
anonymous
  • anonymous
Ah, okey :) So how do I get the argument from that?
anonymous
  • anonymous
Just to try -sqrt3-i and sqrt3-i ?
anonymous
  • anonymous
But they are not included in the interval so instead should i take some random x value inside the interval?
experimentX
  • experimentX
yeah you can get that ... |x| must be sqrt(3)'s ...just put it up and see what you get.
anonymous
  • anonymous
So i should use -i as Im, right?
anonymous
  • anonymous
If so the argument is \[\frac{ 11\pi }{ 6 }\]
experimentX
  • experimentX
put z = x + i in that equation |x + i - 2i| < 2 x^2 + 1^2 < 4 x^2 < 3 |x| < sqrt(3) -sqrt(3) < x < sqrt(3) put z = x + 11pi/6 i <--- in above
anonymous
  • anonymous
\[z=x+\frac{ 11\pi }{6 }i, \] where \[-\sqrt{3}
anonymous
  • anonymous
Is that what you meant?
experimentX
  • experimentX
no ... find the value of x in similar fashion
anonymous
  • anonymous
Sorry but i have to go for now. But thank you for your help and patience, will look back at this later.

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