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Thread: Complex numbers

  1. #1
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    Complex numbers

    Please help me to solve:

    Prove
    $\displaystyle Rez>0 \; \: and \; \; z\neq 1 \Rightarrow \left |\frac{z+1}{z-1} \right |>1$

    Thanks a lot
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  2. #2
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    Quote Originally Posted by sinichko View Post
    Prove
    $\displaystyle Rez>0 \; \: and \; \; z\neq 1 \Rightarrow \left |\frac{z+1}{z-1} \right |>1$
    $\displaystyle \left| z \right|^2 + 2\text{Re} (z) + 1 > \left| z \right|^2 - 2\text{Re} (z) + 1~~~[1]$
    From the given it should be clear that [1] is true.

    Recall that $\displaystyle z + \overline z = 2\text{Re} (z)$

    You need to verify that $\displaystyle \left| z \right|^2 + 2\text{Re} (z) + 1 = \left| z \right|^2 + z + \overline z + 1 = \left( {z + 1} \right)\left( {\overline z + 1} \right)$
    and $\displaystyle \left| z \right|^2 - 2\text{Re} (z) + 1 = \left| z \right|^2 - z - \overline z - 1 = \left( {z - 1} \right)\left( {\overline z - 1} \right)$

    But $\displaystyle \left( {z + 1} \right)\left( {\overline z + 1} \right) = \left| {z + 1} \right|^2 \;\& \,\left( {z - 1} \right)\left( {\overline z - 1} \right) = \left| {z - 1} \right|^2 $ which implies $\displaystyle \left| {z + 1} \right|^2 > \left| {z - 1} \right|^2 $.

    You can finish.
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