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Math Help - Complex function help

  1. #1
    Senior Member Danneedshelp's Avatar
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    Complex function help

    Q: Let w=f(z)=i(\frac{1-z}{1+z}). Show that f maps the open unit disk \{z\in{\mathbb{C}}:|z|<1\} into the upper half-plane \{w\in{\mathbb{C}}:Im(w)>0\}, and maps the unit circle \{z\in{\mathbb{C}}:|z|=1\} to the real line.

    Im am not sure how to approach this. I found a problem similar to this one in my text book. Here is what I have worked out so far:

    Im(f(z))=Im(i(\frac{1-z}{1+z}))=Im(i(\frac{|1-z|^{2}}{(1+z)(1-\bar{z})}))=Im(\frac{|1-z|^{2}i}{1-|z|^{2}})=\frac{|1-z|^{2}i}{1-|z|^{2}}

    Since |z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0<  1-|z|^{2} and |1-z|^{2}>0 for all z, f maps from the open disk of raduis 1 into the upper half plane.

    I am not sure if I interpreting what I read correctly, so any help would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let w=f(z)=i(\frac{1-z}{1+z}). Show that f maps the open unit disk \{z\in{\mathbb{C}}:|z|<1\} into the upper half-plane \{w\in{\mathbb{C}}:Im(w)>0\}, and maps the unit circle \{z\in{\mathbb{C}}:|z|=1\} to the real line.

    Im am not sure how to approach this. I found a problem similar to this one in my text book. Here is what I have worked out so far:

    Im(f(z))=Im(i(\frac{1-z}{1+z}))=Im(i(\frac{|1-z|^{2}}{(1+z)(1-\bar{z})}))=Im(\frac{|1-z|^{2}i}{1-|z|^{2}})=\frac{|1-z|^{2}i}{1-|z|^{2}}

    Since |z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0<  1-|z|^{2} and |1-z|^{2}>0 for all z, f maps from the open disk of raduis 1 into the upper half plane.

    I am not sure if I interpreting what I read correctly, so any help would be appreciated. Thanks.


    Check carefully the denominator in the third equality of your "proof": do you think that (1+z)(1+\overline{z})=1-|z|^2?? I don't think so.
    A similar mistake happened with the numerator in the second expression...

    Tonio
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  3. #3
    Senior Member Danneedshelp's Avatar
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    what would be the correct approach? Should I write out z in the form x+iy and expand everyting out, distribute the i, and group?
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  4. #4
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    You have made a mistake in the fractions: \dfrac{(1-z)}{(1+z)}\dfrac{(1+\overline{z})}{(1+\overline{z}  )} .
    Now (1-z)(1+\overline{z})=1-2i\text{Im}(z)-|z|^2 .
    And (1+z)(1+\overline{z})=|1+z|^2


    So,  \displaystyle \text{Im}(i~f(z))=\dfrac{1-|z|^2}{|1+z|^2}.
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    You have made a mistake in the fractions: \dfrac{(1-z)}{(1+z)}\dfrac{(1+\overline{z})}{(1+\overline{z}  )} .
    Now (1-z)(1+\overline{z})=1-2i\text{Im}(z)-|z|^2 .
    And (1+z)(1+\overline{z})=|1+z|^2


    So,  \displaystyle \text{Im}(i~f(z))=\dfrac{1-|z|^2}{|1+z|^2}.
    So, just to be clear, the conjugate of (1+z) is (1-\bar{z}) and the conjugate of (1-z) is (1+\bar{z})? If so, this means I alwars change the sign and take the conjugate of the complex number in the expression. Correct?

    Now, since |z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0<  1-|z|^{2} and |1+z|^{2}>0 for all z in the domain, it follows that f maps fromt he unit disk into the upper half-plane.

    Is this a sufficient answer to the question?

    Thanks for the help
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  6. #6
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    Quote Originally Posted by Danneedshelp View Post
    So, just to be clear, the conjugate of [tex](1+z)[\math] is (1-\bar{z}) and the conjugate of (1-z) is (1+\bar{z})?
    That is completely wrong.
    First. To rationalize a complex fraction multiply by the conjugate of the denominator.

    So \dfrac{z}{w}=\dfrac{z\overline{w}}{|w|^2}

    Moreover,  \overline{1+z}=1+\bar{z} AND  \overline{1-z}=1-\bar{z}
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