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Math Help - Two equations In : C

  1. #1
    Super Member dhiab's Avatar
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    Two equations In : C

    Solve in C :
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dhiab View Post
    Solve in C :

    {z^4} + {\left( {1 + z} \right)^4} = 0 \Leftrightarrow

    \Leftrightarrow {z^4} + 2{z^2}{\left( {1 + z} \right)^2} + {\left( {1 + z} \right)^4} - 2{z^2}{\left( {1 + z} \right)^2} = 0 \Leftrightarrow

    \Leftrightarrow {\left( {{z^2} + {{\left( {1 + z} \right)}^2}} \right)^2} - 2{z^2}{\left( {1 + z} \right)^2} = 0 \Leftrightarrow

    \Leftrightarrow \left( {{z^2} + {{\left( {1 + z} \right)}^2} - \sqrt 2 z\left( {1 + z} \right)} \right)\left( {{z^2} + {{\left( {1 + z} \right)}^2} + \sqrt 2 z\left( {1 + z} \right)} \right) = 0.

    What you need to do now, I hope you know
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  3. #3
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    Another method: z^4 + (1 + z)^4 = 0\: \Leftrightarrow\: (1+z)^4 = (-1)z^4\: \Leftrightarrow\: 1+z = \omega z \: \Leftrightarrow\: z = \frac1{\omega-1}, where \omega is one of the fourth roots of 1, namely \tfrac1{\sqrt2}(\pm1\pm i).
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by Opalg View Post
    Another method: z^4 + (1 + z)^4 = 0\: \Leftrightarrow\: (1+z)^4 = (-1)z^4\: \Leftrightarrow\: 1+z = \omega z \: \Leftrightarrow\: z = \frac1{\omega-1}, where \omega is one of the fourth roots of 1, namely \tfrac1{\sqrt2}(\pm1\pm i).
    Hello thank you we have anathor solution :
    I can be writeen :
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