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Thread: Finding sum of complex numbers

  1. #1
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    Finding sum of complex numbers

    Find the sum of $\displaystyle 1 + i + i^2 + i^3+...+i^{16}$

    Attempt:

    $\displaystyle \sum_{k=0}^{n}ar^k$
    $\displaystyle =\frac{a(1-r^{n+1})}{1-r}$
    $\displaystyle =\frac{(1-i^{16+1})}{1-r}$
    $\displaystyle =\frac{(1-i^{17})}{1-r}$

    but the correct answer was 1
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  2. #2
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    I get 1. Note that $\displaystyle i^{17}=i$ & $\displaystyle \dfrac{1-i}{1-i}=1$
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  3. #3
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    Quote Originally Posted by Plato View Post
    I get 1. Note that $\displaystyle i^{17}=i$ & $\displaystyle \dfrac{1-i}{1-i}=1$
    How did you get this?
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  4. #4
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    Quote Originally Posted by SyNtHeSiS View Post
    How did you get this?
    $\displaystyle \dfrac{{a\left( {1 - r^{n + 1} } \right)}}
    {{1 - r}},\;a = 1,\;r = i\;\& \;n = 16$

    $\displaystyle \dfrac{{\left( {1 - i^{17} } \right)}}{{1 - i}} = \dfrac{{\left( {1 - i} \right)}}{{1 - i}}$
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  5. #5
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    Quote Originally Posted by SyNtHeSiS View Post
    How did you get this?
    If you are wondering about $\displaystyle i^{17}$

    $\displaystyle i^{16}=\left(i^2\right)^8=(-1)^{even}=1$
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  6. #6
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    I prefer this idea. $\displaystyle \bmod (N,4) = k$ means that $\displaystyle k$ is the remainder when $\displaystyle N$ is divided by $\displaystyle 4$.

    So $\displaystyle i^N=i^{\bmod (N,4)}$

    Example: $\displaystyle i^{17}=i^{\bmod (17,4)}=i^1$
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  7. #7
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    We might also point out that $\displaystyle (1+ x+ x^2+ \cdot\cdot\cdot+ x^{n-1}+ x^n)(1- x)= (1- x)^{n+1}$ for any positive integer n.
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