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frx

  • 2 years ago

Draw the set S of complex numbers such that \[|z-2i|<2,,, \Im (z)=1\] What is the argument when \[z \in S\]

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  1. frx
    • 2 years ago
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    \[|a-i|<2\]

  2. frx
    • 2 years ago
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    But what does that say on a graph?

  3. frx
    • 2 years ago
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    -2<a-i<2

  4. experimentX
    • 2 years ago
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    says it lines in the region inside a circle with radius 2 and center 0,2

  5. frx
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=|a-i|%3C2

  6. frx
    • 2 years ago
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    It's no circle?

  7. frx
    • 2 years ago
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    Is it because the Im(z) is constant at 1 and -2< Re(z) < 2, so it just varies in the Re(z) plane?

  8. experimentX
    • 2 years ago
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    so you want z such that it's imaginary part is constant?

  9. frx
    • 2 years ago
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    No i think I'm a bit wrong in my thoughts

  10. experimentX
    • 2 years ago
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    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

  11. frx
    • 2 years ago
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    Is it not Re betweem -2 and 2 ?

  12. experimentX
    • 2 years ago
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    sorry ... sqrt(3) .. lol

  13. frx
    • 2 years ago
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    Ah, okey :) So how do I get the argument from that?

  14. frx
    • 2 years ago
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    Just to try -sqrt3-i and sqrt3-i ?

  15. frx
    • 2 years ago
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    But they are not included in the interval so instead should i take some random x value inside the interval?

  16. experimentX
    • 2 years ago
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    yeah you can get that ... |x| must be sqrt(3)'s ...just put it up and see what you get.

  17. frx
    • 2 years ago
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    So i should use -i as Im, right?

  18. frx
    • 2 years ago
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    If so the argument is \[\frac{ 11\pi }{ 6 }\]

  19. experimentX
    • 2 years ago
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    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

  20. frx
    • 2 years ago
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    \[z=x+\frac{ 11\pi }{6 }i, \] where \[-\sqrt{3}<x<\sqrt{3}\]

  21. frx
    • 2 years ago
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    Is that what you meant?

  22. experimentX
    • 2 years ago
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    no ... find the value of x in similar fashion

  23. frx
    • 2 years ago
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    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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