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Math Help - sketching complex

  1. #1
    Junior Member
    Joined
    Nov 2009
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    73

    sketching complex

    Sketch the region onto which the sector 0<r<=2/3 and
    0<= theta <= pie/4 is mapped by the transformations

    1) f(z)=1/z

    2)z^3


    where z=re^(itheta)


    I dont understand graphing in the complex plane.
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  2. #2
    Super Member
    Joined
    Aug 2008
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    I'll do the first one. You have f(z)=1/z=\frac{1}{r}e^{-it} right? Gotta' get that far first ok. So suppose I let z=1 e^{it} and let t go from 0 to 2pi. Then f(1 e^{it}) is just a circle of radius one. Suppose I let z=1/4 e^{it}. Ok, then f(1/4 e^{it}) is then a circle of radius 4. Suppose I let z=1/4 e^{it} but this time I only let t go from 0 to pi. Then that's half a circle but it's the lower half because of the negative exponent. How about if I let z=1/4 e^{it} and let t go from 0 to \pi/4 then f(1/4 e^{it}) is 1/8 of a circle of radius 4 in the fourth quadrant going from 0 to -pi/4 and the smaller I let r, the larger the radius of that circle under f(z) because it's 1/r. Ok, what about all of these 1/8 circles with r going from 0 to 2/3? Wouldn't that be the pi/4 sector in the fourth quadrant with inner radius 3/2 but extending to infinity? Like the plot below which shows the region 0<r<2/3 and 0<=t<=pi/4 on the left, and the image of that region under f(z)=1/z on the right except the red part extends out to infinity and I just chopped it.
    Attached Thumbnails Attached Thumbnails sketching complex-mapregion2.jpg  
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