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May 23rd, 2018, 03:00 AM   #1
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Question 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...
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May 23rd, 2018, 03:17 AM   #2
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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.
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May 23rd, 2018, 03:21 AM   #3
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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 05:45 AM.
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May 23rd, 2018, 03:31 AM   #4
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Quote:
Originally Posted by keylewer View Post
I have either proved that 1=2, 0/x for a value of x not equal to 0...
Or I have made a mistake.
This is great, thank you. Too many are unwilling to admit that they might have erred.
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May 23rd, 2018, 01:37 PM   #5
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A simple way to look at it is $e^{2\pi i}=1=e^0$, but $2\pi i\ne 0$.
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