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Math Help - Complex numbers, set of points satisfying an equality

  1. #1
    MHF Contributor arbolis's Avatar
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    Complex numbers, set of points satisfying an equality

    Find the set of points satisfying the equality \left | \frac{z-1}{z+1}  \right |=c, with c\geq 0 an arbitrary constant. (I guess in \mathbb{R}).

    My attempt :
    If c=0 then z=1, which makes a point on the complex plane.
    If c=1 then x=0 and y can be any number in \mathbb{R} so we've a straight line in the complex plane ( x=0) and I believe it will be the same for any other c \neq 0 \in \mathbb{R}.
    Am I right?
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Find the set of points satisfying the equality \left | \frac{z-1}{z+1}  \right |=c, with c\geq 0 an arbitrary constant. (I guess in \mathbb{R}).

    My attempt :
    If c=0 then z=1, which makes a point on the complex plane.
    If c=1 then x=0 and y can be any number in \mathbb{R} so we've a straight line in the complex plane ( x=0) and I believe it will be the same for any other c \neq 0 \in \mathbb{R}.
    Am I right?
    \left | \frac{z-1}{z+1} \right |=c \Rightarrow \left | z-1 \right |=c\left | z+1 \right | \Rightarrow \left | z-1 \right |^2=c^2 \left | z+1 \right |^2 \Rightarrow (z-1)(\bar{z}-1)=c^2(z+1)(\bar{z}+1)\Rightarrow (c^2-1)z\bar{z}+(c^2+1)(z+\bar{z})+(c^2-1)=0

    - If c=0, is the point z=1.
    If c=1, z+\bar{z}=2x=0, is the imaginary axis.
    Otherwise, z\bar{z}+\frac{c^2+1}{c^2-1}(z+\bar{z})+1=0 \Rightarrow \left | z+\frac{c^2+1}{c^2-1} \right | = \frac{2c}{|c^2-1|}, is a circle.
    Last edited by mr fantastic; September 18th 2009 at 08:24 AM. Reason: Restored original reply
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  3. #3
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    Quote Originally Posted by luobo View Post
    \left | \frac{z-1}{z+1} \right |=c \Rightarrow \left | z-1 \right |=c\left | z+1 \right | \Rightarrow \left | z-1 \right |^2=c^2 \left | z+1 \right |^2 \Rightarrow (z-1)(\bar{z}-1)=c^2(z+1)(\bar{z}+1)\Rightarrow (c^2-1)z\bar{z}+(c^2+1)(z+\bar{z})+(c^2-1)=0

    If c=0, is the point z=1.
    If c=1, z+\bar{z}=2x=0, is the imaginary axis.
    Otherwise, z\bar{z}+\frac{c^2+1}{c^2-1}(z+\bar{z})+1=0 \Rightarrow \left | z+\frac{c^2+1}{c^2-1} \right | = \frac{2c}{|c^2-1|}, is a circle.
    In fact, circles such as these have a special name: Circles of Apollonius.

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