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Math Help - 1 sum. elementary transformation from z-plane to w-plane (complex number)

  1. #1
    Member ssadi's Avatar
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    1 sum. elementary transformation from z-plane to w-plane (complex number)

    For the transformation w=z^2 show that as z moves once round the circle centre O and radius 2, w moves twice round the circle center O and radius 4.

    I have two problems for which I am stuck on the sum:

    1. How do moving "once" and "twice" is represented on the equation? That is, if w lies on circle center O and radius 4,  |w|=4 , but how do show that it moves twice?

    2. If i take |z|=2, w=z^2,  z=\sqrt w, |z|=|\sqrt w|=2: how do I do this part: 2=|\sqrt w|, when w is a complex number?
    And how do i carry out from there :help:
    Last edited by ssadi; December 3rd 2008 at 04:10 AM.
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  2. #2
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    Quote Originally Posted by ssadi View Post
    For the transformation w=z^2 show that as z moves once round the circle centre O and radius 2, w moves twice round the circle center O and radius 4.

    I have two problems for which I am stuck on the sum:

    1. How do moving "once" and "twice" is represented on the equation? That is, if w lies on circle center O and radius 4,  |w|=4 , but how do show that it moves twice?

    2. If i take |z|=2, w=z^2,  z=\sqrt w, |z|=|\sqrt w|=2: how do I do this part: 2=|\sqrt w|, when w is a complex number?
    And how do i carry out from there :help:
    The path that you take on the circle at origin of radius two can be expressed as g(\theta) = 2e^{i\theta} for 0\leq \theta \leq 2\pi.
    Under the transformation z\mapsto z^2 the path becomes mapped to \left( 2e^{i\theta} \right)^2 = 4e^{2i\theta} for 0\leq \theta \leq 2\pi. Because of the presence of 2i\theta (rather than i\theta) it means the points moves twice around a circle of radius 4.
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  3. #3
    Member ssadi's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The path that you take on the circle at origin of radius two can be expressed as g(\theta) = 2e^{i\theta} for 0\leq \theta \leq 2\pi.
    Under the transformation z\mapsto z^2 the path becomes mapped to \left( 2e^{i\theta} \right)^2 = 4e^{2i\theta} for 0\leq \theta \leq 2\pi. Because of the presence of 2i\theta (rather than i\theta) it means the points moves twice around a circle of radius 4.
    Gotcha, thanks
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