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SheldonEinstein Group Title

Let z be a complex number than locus represented by |iz-1| + |z-i| = 2 is : a) a line , b) a circle , c) a pair of straight lines, d) a parabola

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    i think it is a l ine

    • 2 years ago
  2. SheldonEinstein Group Title
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    Sorry for the mistake in the question earlier , the modified quest. is in the post itself. I seek help from the users presented here as soon as possible,

    • 2 years ago
  3. SheldonEinstein Group Title
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    @satellite73 rethink please :)

    • 2 years ago
  4. satellite73 Group Title
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    scratch that, i think it is a region bounded by two line

    • 2 years ago
  5. SheldonEinstein Group Title
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    OK so since it is IIT based so I will say : There may be no answers or multiple answers

    • 2 years ago
  6. SheldonEinstein Group Title
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    Well yes @satellite73 has the answer, it will be a line \(\textbf{segment}\)

    • 2 years ago
  7. SheldonEinstein Group Title
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    But I am half way stuck @satellite73 , can you show your work, I think @experimentX is also writing his work :)

    • 2 years ago
  8. experimentX Group Title
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    put z = x + iy you would get, \[ \sqrt{(y-1)^2 + x^2} + \sqrt{x^2 + (y-1)^2} = 2\] http://www.wolframalpha.com/input/?i=plot+\sqrt{%28y-1%29^2+%2B+x^2}+%2B+\sqrt{x^2+%2B+%28y-1%29^2}+%3D+2 guess it is a circle

    • 2 years ago
  9. experimentX Group Title
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    woops!! (y+1)^2

    • 2 years ago
  10. SheldonEinstein Group Title
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    What I can do is : \[\large{|(-iz-1)| + |(z-i)| =2}\] \[\large{|i(z+i)|+|z-i|=2}\] \[\large{|i||z+i| + |z-i| = 2 }\] \[\large{|z+i| + |z-i| = 2}\]

    • 2 years ago
  11. satellite73 Group Title
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    that looks good

    • 2 years ago
  12. SheldonEinstein Group Title
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    |dw:1352908224557:dw|

    • 2 years ago
  13. SheldonEinstein Group Title
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    So am I right in my way @experimentX @satellite73 ? (Since I am only given 54 seconds to do these type of questions so I prefer to use shortcuts that is here : not putting z = x + iy , my opinion said , yes ! x will be equal to zero if we are going to solve it! )

    • 2 years ago
  14. satellite73 Group Title
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    yes \(x=0\)

    • 2 years ago
  15. satellite73 Group Title
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    i did it the donkey way but your way is much much better i think

    • 2 years ago
  16. UnkleRhaukus Group Title
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    \(z=i\) is a solution

    • 2 years ago
  17. SheldonEinstein Group Title
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    Oh! No problem, sometimes we learn a lot from "that say donkey ways" , thanks for your time : @experimentX @satellite73

    • 2 years ago
  18. experimentX Group Title
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    |z| <= 1 and z = i

    • 2 years ago
  19. experimentX Group Title
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    no wonder it does not show in x's and y's

    • 2 years ago
  20. UnkleRhaukus Group Title
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    \(z=-i\) is also a solution

    • 2 years ago
  21. experimentX Group Title
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    the solution is a line segment from +i to -i

    • 2 years ago
  22. UnkleRhaukus Group Title
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    \[\large{|z+i| + |z-i| = 2}\] change of variables shift along the complex plane \[w=z+i\] \[{|w| + |w-2i| = 2}\]

    • 2 years ago
  23. UnkleRhaukus Group Title
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    |dw:1352909335544:dw|

    • 2 years ago
  24. UnkleRhaukus Group Title
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    circle + circle = circle?

    • 2 years ago
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