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Math Help - Logarithm Complex Number Help!

  1. #1
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    Logarithm Complex Number Help!

    If i = sqrt(-1), find the value of the sum ilog1 + i^2log2 + ... + i^nlog(n) + ... + (i^10)log10
    Last edited by Espionage; June 19th 2013 at 03:13 PM.
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  2. #2
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    Re: Logarithm Complex Number Help!

    Quote Originally Posted by Espionage View Post
    If i = sqrt(-1), find the value of the sum ilog1 + i^2log2 + ... + i^nlog(n) + ... + (i^10)log10
    Have a look at this.

    Because \log(1)=0 we start the sum index with k=2
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  3. #3
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    Re: Logarithm Complex Number Help!

    i^2= -1, i^3= -i, and i^4= 1. That is, i^n is "periodic" with period 4. Your sum is
    -log(2)- ilog(3)+ log(4)+ i log(5)- log(6)- ilog(7)+ log(8)+ i log(9)- log(10)= -(log(2)- log(4)- log(6)+ log(8)- log(10))- i(log(3)- log(5)- log(7)+ log(9))
    =  log\left(\frac{32}{240}\right)- i log\left(\frac{27}{35}\right)= log\left(\frac{2}{15}\right)- i log\left(\frac{27}{35}\right).
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    Re: Logarithm Complex Number Help!

    Quote Originally Posted by HallsofIvy View Post
    i^2= -1, i^3= -i, and i^4= 1. That is, i^n is "periodic" with period 4. Your sum is
    -log(2)- ilog(3)+ log(4)+ i log(5)- log(6)- ilog(7)+ log(8)+ i log(9)- log(10)= -(log(2)- log(4)- log(6)+ log(8)- log(10))- i(log(3)- log(5)- log(7)+ log(9))
    =  log\left(\frac{32}{240}\right)- i log\left(\frac{27}{35}\right)= log\left(\frac{2}{15}\right)- i log\left(\frac{27}{35}\right).
    Compare this reply with the wolframalpha's result.

    Note that \log \left( {\frac{{15}}{2}} \right) =  - \log \left( {\frac{2}{{15}}} \right).

    But I have no idea where \log\left(\frac{27}{35}\right) comes from?

















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