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May 23rd, 2018, 04:00 AM  #1 
Newbie Joined: May 2018 From: United Kingdom Posts: 1 Thanks: 1  1=2 or 0/x NOT = 0
I have either proved that 1=2, 0/x for a value of x not equal to 0... Or I have made a mistake. I started with e^i*pi=1 I then squared both sides to get: e^2*i*pi=1 Taking the log of both sides allows me to remove the power and multiply by the log for: 2*i*pi*log(e)=log(1) S1) 2*i*pi*log(e)=0 To extract the value of i, sqrt(1), I divide both sides by 2*pi*log(e) i=0/2*i*pi*log(e) 0/x=0 i=0 sqrt(1)=0 Square both sides for: 1=0 +2 both sides: 1=2 Help... 
May 23rd, 2018, 04:17 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond 
Your mistake lies in applying logarithmic identities which apply to real numbers to a complex number. Not all of these identities are necessarily true for complex numbers. See here for more information. 
May 23rd, 2018, 04:21 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
Your error is when you take the logarithm of a complex number. A complex number can be written in polar form as $\displaystyle re^{i\theta}= re^{i(\theta+ 2\pi n)}$ since "$\displaystyle e^{i\theta}= \cos(\theta)+ i\sin(\theta)$" is periodic with period $\displaystyle 2\pi$. Taking the logarithm, $\displaystyle \ln(re^{i(\theta+ 2n\pi)}= \ln(r)+ i(\theta+ 2n\pi)$ for n any integer. That is, ln is multivalued. Last edited by skipjack; May 23rd, 2018 at 06:45 AM. 
May 23rd, 2018, 04:31 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,557 Thanks: 2558 Math Focus: Mainly analysis and algebra  
May 23rd, 2018, 02:37 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,660 Thanks: 648 
A simple way to look at it is $e^{2\pi i}=1=e^0$, but $2\pi i\ne 0$.
