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Math Help - Complex number question

  1. #1
    Senior Member Stroodle's Avatar
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    Complex number question

    Is there a relatively fast way of solving this?

    If f(i)=\frac{1+i+i^2+...+i^{11}}{1-i} which of the following statements are true?

    (a) f(i)=2+i

    (b) Re[f(i)]=5

    (c) Im[f(i)]=8

    (d) f(i)=1-i

    (e) f(i)=0

    Thanks for your help!
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  2. #2
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    You can simplify the equation by using the fact that :

    i^2 = -1
    i^3 = (i)(i^2)=-i
    i^4 = (i^2)^2 = 1
    i^5 = (i^4) (i)=i
    and so on
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  3. #3
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    You can multiply the f(i) by 1+i/1+i to get rid of the complex number in the denominator..
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  4. #4
    Senior Member Stroodle's Avatar
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    Yep. But I was wondering if there's a faster way to do it.
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  5. #5
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    How about muliplying by i-1?
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  6. #6
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    songoku's method is the fastest. You need to recognize the fact that the powers of i cycle through four different numbers: 1, i, -1, and -i. The sum of any four consecutive powers of i will be 0. The fraction could be written like this:

    \frac{i^0 + i^1 + i^2 + i^3 + ... + i^{11}}{i - 1}

    The sum of the first four powers of i shown above is

    i^0 + i^1 + i^2 + i^3 = 1 + i - 1 - i = 0

    So how many powers of i are being added all together? What is the sum in the numerator? If you can answer these questions, you'll know what the fraction simplifies to.


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