I was playing around with euler's formula the other day and found this odd proof which says 2ipi=0. I know this is obviously wrong, but what was the step that went wrong? The "proof" goes like this:(adsbygoogle = window.adsbygoogle || []).push({});

start with e^(i*pi)+1=0

so -e^(i*pi)=1

multiply both sides by e e(-e^(i*pi))=e

factor a (-1) (-1)(e)(e^(i*pi))=e

add exponents (-1)(e^(i*pi+1))=e

take ln of both sides ln((-1)(e^(i*pi+1)))=1

use properties of logs ln(-1) + ln(e^(i*pi+1))=1

take ln of -1 i*pi + ln(e^(i*pi+1))=1

properties of logs i*pi + (i*pi+1)ln(e)=1

take ln(e) i*pi + i*pi+1=1

subtract 1 i*pi + i*pi=0

combine like terms 2*pi*i=0 ...

this even works in the original formula:

e^(i*2*pi)=cos(2*pi)+isin(2*pi)

e^(i*2*pi)=1

e^(i*2*pi)=e^0

i*2*pi=0 ...

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# 2*pi*i = 0? euler's formula

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