Another fallacy

  • Thread starter mathelord
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mathelord

HI, since e^[i*pi]=-1,then e^[2i*pi]=1.
taking lin of both sides,2*i*pi=0,that makes i=0
squaring both sides,you get -1 to equal 0.
Is there a problem with what i have done,and what is it?
 

Curious3141

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When complex outputs are allowed, logarithms become multivalued. In this case, the correct formulae are :

[tex]e^{2k\pi i} = 1, \ln(1) = 2k\pi i[/tex] where [itex]k[/itex] ranges over the integers.

In your case, you're confusing the result for k = 0 with k = 1, hence the apparent paradox.
 
Last edited:

Zurtex

Science Advisor
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Yes and:

[tex](-1)^2 = 1 \quad 1^2 = 1 \quad 1 = - 1 \quad 2 = 0[/tex]

It's the same things really :rolleyes:
 

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