# Logarithm Complex Number Help!

• Jun 19th 2013, 04:10 PM
Espionage
Logarithm Complex Number Help!
If i = $sqrt(-1)$, find the value of the sum $ilog1 + i^2log2 + ... + i^nlog(n) + ... + (i^10)log10$
• Jun 19th 2013, 04:27 PM
Plato
Re: Logarithm Complex Number Help!
Quote:

Originally Posted by Espionage
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 $k=2$
• Jun 19th 2013, 05:38 PM
HallsofIvy
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))
$= log\left(\frac{32}{240}\right)- i log\left(\frac{27}{35}\right)= log\left(\frac{2}{15}\right)- i log\left(\frac{27}{35}\right)$.
• Jun 19th 2013, 06:26 PM
Plato
Re: Logarithm Complex Number Help!
Quote:

Originally Posted by HallsofIvy
$i^2= -1$, $i^3= -i$, 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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