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Math Help - Complex Conjugate

  1. #1
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    Complex Conjugate

    Let (z) be the complex conjugate of z.

    Here's the problem:

    Let c e C be non-zero(c is a non-zero element of the complex numbers)
    Show (z)^-1 = (z^-1)
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  2. #2
    MHF Contributor red_dog's Avatar
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    (\overline{z})^{-1}=\frac{1}{\overline{z}}

    \overline{z^{-1}}=\overline{\left(\frac{1}{z}\right)}=\frac{\ove  rline{1}}{\overline{z}}=\frac{1}{\overline{z}}
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  3. #3
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    Alternatively, using DeMoivre's Theorem...

    z = r(\cos{\theta} + i\sin{\theta}), \overline{z} = r(\cos{\theta} - i\sin{\theta}).

    z^{-1} = r^{-1}[\cos{(-\theta)} + i\sin{(-\theta)}]

     = \frac{1}{r}(\cos{\theta} - i\sin{\theta}).

    So \overline{z^{-1}} = \frac{1}{r}(\cos{\theta} + i\sin{\theta}).



    (\overline{z})^{-1} = r^{-1}[\cos{(-\theta)} - i\sin{(-\theta)}]

     = \frac{1}{r}(\cos{\theta} + i\sin{\theta})

     = \overline{z^{-1}}.



    Or using Cartesians:

    z = a+ib, \overline{z} = a-ib


    z^{-1} = \frac{1}{z}

     = \frac{1}{a + ib}

     = \frac{a - ib}{a^2 + b^2}

     = \frac{a}{a^2 + b^2} -i\left(\frac{b}{a^2 + b^2}\right)


    \overline{z^{-1}} = \frac{a}{a^2 + b^2} + i\left(\frac{b}{a^2 + b^2}\right).



    (\overline{z})^{-1} = \frac{1}{\overline{z}}

     = \frac{1}{a - ib}

     = \frac{a + ib}{a^2 + b^2}

     = \frac{a}{a^2 + b^2} + i\left(\frac{b}{a^2 + b^2}\right)

     = \overline{z^{-1}}.
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