Is 2 pi i equal to 0, contradicting the fact that pi and i cannot equal 0?

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In summary, the conversation discusses the use of logarithms and the definition of LN in the complex plane. It is important to note that logs of complex numbers are only defined up to multiples of 2pi and the mistake lies in assuming that ln(e^z) is equal to z for complex numbers. The correct statement is exp(ln(z)) = z, where z is not equal to 0.
  • #1
nicktacik
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This one has me stumped.

[tex]e^{\pi i} = -1[/tex]
[tex]e^{2 \pi i} = (-1) ^ 2 = 1 [/tex]
[tex]ln(e^{2 \pi i}) = ln(1) = 0 [/tex]
[tex]2 \pi i = 0 [/tex]

Or is 2 pi i actually 0, and this does not actually imply that either pi = 0 or i = 0?
 
Last edited:
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  • #2
Since [itex] 2 \pi i [/itex] is an imaginary number you must use the definition of LN in the complex plane. You are using the definition for the real number line.

in the complex plane we have:

[tex] logz = log(r) + i \theta [/tex]

edit (changed my definition of z)
[tex] r = |z| [/tex]
and
[tex] \theta = arg(z) [/tex]
 
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  • #3
More than that - logs of complex numbers are only defined up to multiples of 2pi. One can ought to take the principal branch - this is just a slightly more complicated variation on the square root 'fallacies'.
 
  • #4
The mistake is here:
[tex] \ln e^z = z[/tex]
This is not true for complex numbers.

Note: the other way around:
[tex] \exp (\ln z) = z, \ z\not =0[/tex]
Is true.
 
  • #5
Ok thanks. As you can see, I haven't taken my complex variables class yet.
 
  • #6
its sort of like saying, (2^2 =4 and (-2)^2 = 4 so 2 = -2.)
 

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