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Mappings in the complex plane

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data
    What is the mapping of the circle of radius 1 centered at z=-2i
    under the mappinf f(z)=1/z
    3. The attempt at a solution
    I write the circle in polar form [itex] -2i+e^{ix} [/itex]
    Now we invert it and multiply by the complex conjugate.
    so we get [itex] f(z)= \frac{2i+e^{-ix}}{5+2i(e^{ix}-e^{-ix})} [/itex]
    now sure how to graph it our manipulate it any more, should I break it into real and imaginary parts?
    also there was a hint that w=1/z and z=1/w. where w is the complex mapping. any help would be appreciated
     
  2. jcsd
  3. Feb 9, 2017 #2

    BvU

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    Well, we multiply by 1 where we write 1 as cc/cc. The denominator is now real -- perhaps you want to let it look a little more real (think goniometric functions) ? And for the numerator real and imaginary parts are easily separated...

    [edit] I see that there's more to be had here (the image is very simple), but can't pin it down
     
    Last edited: Feb 9, 2017
  4. Feb 9, 2017 #3
    thanks for your help, ok when i plug in sin(x) instead of the complex exponentials I get
    [itex] \frac{i(2-sin(x)}{5-4sin(x)}+\frac{cos(x)}{5-4sin(x)} [/itex]
    Not sure what to do from here, I could plug in points and graph it, but I thought this was to be something somewhat straight forward
    to graph.
     
  5. Feb 9, 2017 #4

    BvU

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    I did just that and it inspired me to the [edit] addition. Just trying to figure it out.
     
  6. Feb 10, 2017 #5
    I guess another way to try and solve it is using the modulus of a circle.
    w=1/z. So the radius is one . So we have [itex] |z+2i|=1=|1/w+2i|=1 [/itex]
    then we get [itex] |\frac{2iw+1}{w}|=1 [/itex]
    then we multiply both sides by |w| and square both sides . Then I get
    [itex] 1+4w^2=x^2+y^2 [/itex]
    then I write w as x+iy, square it ans then equate the real and imaginary parts.
    is this the way to go?
     
  7. Feb 10, 2017 #6

    BvU

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    Don't know what the modulus of a circle is. Plug and plot (plod :smile:) gave me another circle as image ...
     
  8. Feb 10, 2017 #7
    ok but how do I get to the circle with algebra ?
     
  9. Feb 11, 2017 #8
    If ##z=e^{it}## then ##w=\frac{1}{e^{it}}=e^{-it}##. So as ##z## traverses the circle in a counter-clockwise direction, what is ##e^{-it}## doing?
     
  10. Feb 11, 2017 #9
    with the negative sign it goes around the circle in the opposite direction., but when I put the shift in the -2i translation, it makes the algebra more difficult,
    my teacher said using the modulus form of circle to make the algebra more clean and simple, but I cant quite get it to work. thanks for your help by the way.
     
  11. Feb 11, 2017 #10

    BvU

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    Do you already have the equation of the image circle ?
     
  12. Feb 11, 2017 #11
    I got it figured out, i just need to be more careful with the i and w . thanks for your help.
     
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