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Thread: complex numbers

  1. #1
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    complex numbers

    Can someone please tell me why this statement is true:

    $\displaystyle (1+i)^{200} = 2^{100}$

    Here's what I did:

    $\displaystyle ((1+i)^2)^{100} = 2^{100} \rightarrow (1+i)^2 = 2 \rightarrow 2i \neq 2$

    as in it's false... so what's my mistake???
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  2. #2
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    because $\displaystyle i^{100} = (i^4)^{25} = 1^{25} = 1 $

    The equation $\displaystyle x^{100} = 2^{100} $ doesn't mean $\displaystyle x = 2 $ is the only solution , there are other solutions such as $\displaystyle x = -2 , 2i , -2i $ etc .
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  3. #3
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    $\displaystyle (1+i)^2=2i\Longrightarrow (1+i)^{200}=\left((1+i)^2\right)^{100}=(2i)^{100}= 2^{100}i^{100}=2^{100}$ , since the

    powers of $\displaystyle i$ are periodic modulo 4.

    And your mistake was in $\displaystyle a^{100}=b^{100}\Longrightarrow a=b$, since it should be $\displaystyle a^{100}=b^{100}\Longrightarrow \left(\frac{a}{b}\right)^{100}=1\Longrightarrow \frac{a}{b}$ is

    one of the 100 different roots of unit of order 100...or, of course, $\displaystyle a=b=0$

    Tonio
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  4. #4
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    $\displaystyle \left( {1 + i} \right)^{200} = \left( {\sqrt 2 } \right)^{200} \left[ {\cos \left( {\frac{\pi }
    {4}} \right) + i\sin \left( {\frac{\pi }
    {4}} \right)} \right]^{200} $
    $\displaystyle =\left( {\sqrt 2 } \right)^{200} \left[ {\cos \left( {\frac{{200\pi }}
    {4}} \right) + i\sin \left( {\frac{{200\pi }}
    {4}} \right)} \right] = 2^{100} $
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