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Math Help - Image for the region under the transformation w=z^2

  1. #1
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    Image for the region under the transformation w=z^2

    Hi there. I have to find the image under the transformation w=z^2 for the region given in the pic.


    So, I tried to analyze what would happend. I know that:
    w=z^2=|z|^2 e^{i2\theta}=|z|^2(\cos 2\theta+i\sin 2\theta)

    And I've tried to express the region under consideration, I did it this way, I'm not pretty sure if this is right.
    1 \geq Re(z) \geq 0
    1 \geq Im(z) \geq 1-Re(z)
    \sqrt{2} \geq |z| \geq 1/2
    \pi/2 \geq arg(z) \geq 0

    Then I should get
    2 \geq |w| \geq 1/4
    \pi \geq arg(w) \geq 0

    But doesn't give enough information to get the image under the transformation. So then I've tried to get the image for the lines.

    y=1\rightarrow{v=1},\alpha \in{[\pi/2,\pi]}
    u=1\rightarrow{u=1},\alpha \in{[0,\pi/2]}

    Which gives something like semi infinite lines in w, right? but that would be wrong having in mind the limitations for the modulus in w.
    And I don't know how to trasnform the other line, y=1-x.
    Attached Thumbnails Attached Thumbnails Image for the region under the transformation w=z^2-conformal.png  
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  2. #2
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    Re: Image for the region under the transformation w=z^2

    I would do this a little bit differently.

    First let u=\text{Re}(w), v=\text{Im}(x)

    Then z^2=(x+iy)^2=x^2-y^2+2xyi

    This gives that u=x^2-y^2, v=2xy

    Now you can just parameterize each of the line segments.

    The vertical line segment from (1,0) to (1,1) is given by

    x(t)=1, y(t)=t, t \in [0,1]

    Under the transformation this gives

    u=1-t^2, v=2t so the start and end points are (1,0) and (0,2) now if you elminate the paramter you get the curve

    u=1-\left(\frac{v}{2} \right) \iff u=1-\frac{v^2}{4} So this is a parabolic arc.

    Now for the horizontal line segemtn from (1,1) to (0,1) we have

    x=1-t, y=1 under the transformation this gives

    u=(1-t)^2-1^2=t^2-2t, v= 2(1-t)1=-2t+2

    If you eliminate the parameter you get

    u=\frac{1}{4}v^2-1

    we get another parabolic region with start point (0,1) and end point (-1,0)

    Finally the line can be paramterized by

     x=t, y=1-t under the transformation this gives

    u=t^2-(1-t)^2=-1+2t, v=2t(1-t)=-2t^2+2t If we eliminate the parameter we get

    v=-\frac{1}{2}(u^2-1) another parabola with start point (-1,0) and end point (1,0)

    Here is a plot of the transformation. The first curve is in green, the 2nd in red, and the last in blue.
    Attached Thumbnails Attached Thumbnails Image for the region under the transformation w=z^2-region.jpg  
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  3. #3
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    Re: Image for the region under the transformation w=z^2

    Nice. Thank you verymuch
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