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Thread: Complex Mappings

  1. #1
    MHF Contributor
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    Complex Mappings

    According to the Remarks in Section 2.5, since $\displaystyle f(z)=\frac{1}{z}$ is its own inverse function, the mapping $\displaystyle w=\frac{1}{z}$ on the extended complex plane maps the circle $\displaystyle |z-\frac{1}{2}|=\frac{1}{2}$ to the line Re(w) = 1. Verify this.

    I am not sure how to do this without working backwards which isn't an easy method. Does anyone know how this can be done in a more efficient manner?

    Thank you for your time,

    Dustin.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dwsmith View Post
    According to the Remarks in Section 2.5, since $\displaystyle f(z)=\frac{1}{z}$ is its own inverse function, the mapping $\displaystyle w=\frac{1}{z}$ on the extended complex plane maps the circle $\displaystyle |z-\frac{1}{2}|=\frac{1}{2}$ to the line Re(w) = 1. Verify this.

    I am not sure how to do this without working backwards which isn't an easy method. Does anyone know how this can be done in a more efficient manner?

    Thank you for your time,

    Dustin.
    On the circle you may put $\displaystyle z=\frac{1}{2}(1+e^{i\theta}), \theta \in [0,2\pi)$

    Now compute $\displaystyle w=\frac{1}{z}$ and show that its real part is equal to $\displaystyle $$ 1$ and its imaginary part takes every real value for some $\displaystyle $$ \theta \in [0,2\pi)$

    CB
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