# Thread: Logarithm Complex Number Help!

1. ## Logarithm Complex Number Help!

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

2. ## Re: Logarithm Complex Number Help!

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 $\displaystyle k=2$

3. ## 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))
$\displaystyle = log\left(\frac{32}{240}\right)- i log\left(\frac{27}{35}\right)= log\left(\frac{2}{15}\right)- i log\left(\frac{27}{35}\right)$.

4. ## Re: Logarithm Complex Number Help!

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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