# Logarithm Complex Number Help!

• Jun 19th 2013, 03:10 PM
Espionage
Logarithm Complex Number Help!
If i = $\displaystyle sqrt(-1)$, find the value of the sum $\displaystyle ilog1 + i^2log2 + ... + i^nlog(n) + ... + (i^10)log10$
• Jun 19th 2013, 03:27 PM
Plato
Re: Logarithm Complex Number Help!
Quote:

Originally Posted by Espionage
If i = $\displaystyle sqrt(-1)$, find the value of the sum $\displaystyle ilog1 + i^2log2 + ... + i^nlog(n) + ... + (i^10)log10$

Have a look at this.

Because $\displaystyle \log(1)=0$ we start the sum index with $\displaystyle k=2$
• Jun 19th 2013, 04:38 PM
HallsofIvy
Re: Logarithm Complex Number Help!
$\displaystyle i^2= -1$, $\displaystyle i^3= -i$, and $\displaystyle i^4= 1$. That is, $\displaystyle i^n$ 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))
$\displaystyle = log\left(\frac{32}{240}\right)- i log\left(\frac{27}{35}\right)= log\left(\frac{2}{15}\right)- i log\left(\frac{27}{35}\right)$.
• Jun 19th 2013, 05:26 PM
Plato
Re: Logarithm Complex Number Help!
Quote:

Originally Posted by HallsofIvy
$\displaystyle i^2= -1$, $\displaystyle i^3= -i$, and $\displaystyle i^4= 1$. That is, $\displaystyle i^n$ 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))
$\displaystyle = log\left(\frac{32}{240}\right)- i log\left(\frac{27}{35}\right)= log\left(\frac{2}{15}\right)- i log\left(\frac{27}{35}\right)$.

Compare this reply with the wolframalpha's result.

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

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

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