+0  
 
+2
333
4
avatar

Can this identity be proven, and if so how?  π^2/6 = 1/6 (-i log(-1))^2  I thank you.

Guest Nov 27, 2017
 #1
avatar+20011 
+2

Can this identity be proven, and if so how?  π^2/6 = 1/6 (-i log(-1))^2

 

\(\begin{array}{|rcll|} \hline && \frac{1}{6}\cdot \Big(-i \log(-1) \Big)^2 \\ && &\small{ \begin{array}{|rcll|} \hline \log(z) &=& \ln|z| + i\underbrace{\arg(z)}_{=\arctan(\frac{b}{a}) } \quad | \quad z = a+b\cdot i \\ \log(-1) &=& \ln|-1| + i\underbrace{\arg(-1)}_{= \underbrace{\arctan\left(\frac{0}{-1}\right)}_{=-\pi} } \quad | \quad z = -1+0\cdot i \\ \log(-1) &=& \ln|-1| + i\cdot(-\pi) \\ \log(-1) &=& \ln(1) - i\cdot \pi \quad | \quad \ln(1) = 0 \\ \log(-1) &=& 0 - i\cdot \pi \\ \log(-1) &=& - i\cdot \pi \\ \hline \end{array}} \\ &=& \frac{1}{6}\cdot \Big(-i \cdot(- i\cdot \pi) \Big)^2 \\ &=& \frac{1}{6}\cdot (i^2 \cdot \pi )^2 \quad | \quad i^2 = -1 \\ &=& \frac{1}{6}\cdot (-\pi )^2 \\ &=& \frac{1}{6}\cdot \pi ^2 \\ &=& \dfrac{\pi ^2}{6} \\ \hline \end{array}\)

 

laugh

heureka  Nov 28, 2017
 #2
avatar
+1

Brilliant heureka, as usual. Thank you very much.

Guest Nov 28, 2017
 #3
avatar+7023 
+1

\(\because e^{i\pi} = -1\\ \text{We can immediately imply that} \log(-1) = i\pi\\ \dfrac{1}{6}(-i\log(-1))^2 \\ =\dfrac{1}{6}(-i\cdot i\pi)^2\quad\boxed{-i\cdot i = 1}\\ =\dfrac{1}{6}\pi^2\\ \)

Hence, proved.

MaxWong  Nov 28, 2017
 #4
avatar
0

Max: You are simply beyond words !! Bravo and thanks.

Guest Nov 28, 2017

14 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.