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Shifting a 'complex' circle

  1. Mar 10, 2013 #1
    We typically have z=r*e^(i*theta). But let's say I want a circle centered at 2i.
    Is it valid to write z=r*e^(i*theta)+2i ?

    I ask this because I don't want to have abs(z-2i)=r; I want to solve for z.
     
  2. jcsd
  3. Mar 10, 2013 #2

    tiny-tim

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    hi ericm1234! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yes, a circle centre a and radius b is z - a = be, for all values of θ
     
  4. Mar 10, 2013 #3
    Awesome.
     
  5. Mar 12, 2013 #4

    HallsofIvy

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    Note that, taking z= x+ iy, [itex]|z- 2i|= |x+ (y-2)i|= r[/itex] is the same as [itex]\sqrt{x^23+ (y- 2)^2}= r[/itex] so that [itex]x^2+ (y- 2)^2= r^2[/itex], the circle of radius r with center at (0, 2) or 2i in the complex plane.
     
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