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Math Help - complex no. (cosx+cos2x+...+cos2nx+i(sinx+...sin2nx))=0

  1. #1
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    complex no. (cosx+cos2x+...+cos2nx+i(sinx+...sin2nx))=0

    i don't understand this step. sorry!

    complex no. (cosx+cos2x+...+cos2nx+i(sinx+...sin2nx))=0-out0106.jpeg
    complex no. (cosx+cos2x+...+cos2nx+i(sinx+...sin2nx))=0-out0105.jpeg
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  2. #2
    Super Member wingless's Avatar
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    Let a = \cos\theta + i\sin\theta = \text{cis} \theta

    (\cos\theta + i\sin\theta)+(\cos\theta + i\sin\theta)^2+(\cos\theta + i\sin\theta)^3+...+(\cos\theta + i\sin\theta)^{2n}

    a + a^2 + a^2 + ...+ a^{2n} = \sum_{k=1}^{2n} a^k

    Use the identity \sum_{i=m}^{n}x^i = \frac{x^{n+1}-x^m}{x-1} to find the sum.
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  3. #3
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    Hello, afeasfaerw23231233!


    S \;=\;(\cos\theta + i\sin\theta) + (\cos\theta + i\sin\theta)^2 + (\cos\theta + i\sin\theta)^3 + \hdots + (\cos\theta + i\sin\theta)^{2n}

    This is a geometric series . . .
    . . first term: \cos\theta + i\sin\theta)" alt="a \:=\\cos\theta + i\sin\theta)" />
    . . common ratio: \cos\theta + i\sin\theta)" alt="r \:=\\cos\theta + i\sin\theta)" />
    . . number of terms: 2n


    Recall that the sum of a geometric series is: . S_n \:=\:a\,\frac{1-r^n}{1-r}

    So we have: . S \;=\;\overbrace{(\cos\theta+i\sin\theta)}^a \,\frac{\overbrace{1 -(\cos\theta + i\sin\theta)^{2n}}^{1-r^n}}{\underbrace{1 - (\cos\theta + i\sin\theta)}_{1-r}}

    Got it?

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