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Thread: Result of a function (complex numbers)

  1. #1
    Member Pranas's Avatar
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    Result of a function (complex numbers)

    I need to define the possible outcome of a function

    $\displaystyle \displaystyle \[\omega \left( {\rm Z} \right) = \frac{{2i \cdot {\rm Z}}}{{{\rm Z} + 3}}\]$

    whereas

    $\displaystyle \displaystyle \[\omega = u + i \cdot v \in \mathbb{C}\]$

    $\displaystyle \displaystyle \[{\rm Z} = x + i \cdot y \in \mathbb{C}\]$

    and my $\displaystyle \displaystyle \[{\rm Z}\]$ obeys the equality

    $\displaystyle \displaystyle \[\left| {{\rm Z} - 1} \right| = 2\]$

    I find it fairly complicated, because while computer graphs it nicely as $\displaystyle \displaystyle \[\left| {u + i \cdot v} \right| = 1\]$ , my writings (fragment added below) and squared parts of equation were not going to be nearly as easily interpreted.

    Maybe you know a smart and elegant way? It was not supposed to be a very complicated problem...

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  2. #2
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    Quote Originally Posted by Pranas View Post
    I need to define the possible outcome of a function

    $\displaystyle \displaystyle \[\omega \left( {\rm Z} \right) = \frac{{2i \cdot {\rm Z}}}{{{\rm Z} + 3}}\]$

    whereas

    $\displaystyle \displaystyle \[\omega = u + i \cdot v \in \mathbb{C}\]$

    $\displaystyle \displaystyle \[{\rm Z} = x + i \cdot y \in \mathbb{C}\]$

    and my $\displaystyle \displaystyle \[{\rm Z}\]$ obeys the equality

    $\displaystyle \displaystyle \[\left| {{\rm Z} - 1} \right| = 2\]$

    I find it fairly complicated, because while computer graphs it nicely as $\displaystyle \displaystyle \[\left| {u + i \cdot v} \right| = 1\]$ , my writings (fragment added below) and squared parts of equation were not going to be nearly as easily interpreted.

    Maybe you know a smart and elegant way? It was not supposed to be a very complicated problem...

    Start from the third line of your solution: $\displaystyle z = \dfrac{-3w}{w-2i}.$

    Then $\displaystyle z-1 = \dfrac{-4w+2i}{w-2i}$, so $\displaystyle |z-1| = \dfrac{|-4w+2i|}{|w-2i|} = 2.$ Hence $\displaystyle |-2w+i| = |w-2i|.$

    Therefore $\displaystyle |-2w+i|^2 = |w-2i|^2$. Now use the fact that if $\displaystyle \omega = \xi+i\eta$ then $\displaystyle |\omega|^2 = \xi^2 + \eta^2$, to see that $\displaystyle (-2u)^2 + (-2v+1)^2 = u^2 + (v-2)^2$. Simplify that to get the equation of a circle.
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  3. #3
    Member Pranas's Avatar
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    Thank you! That's exactly what I was looking for
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    BTW the answer is particularly simple.
    Spoiler:
    It reduces to $\displaystyle |w| = 1$.
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