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

  1. Jan 8, 2009 #1
    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:

    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 ...
     
  2. jcsd
  3. Jan 9, 2009 #2
    Re: 2*pi*i=0?

    Well what's the period of sine and cosine? What's e^(4pi*i)?
     
  4. Jan 9, 2009 #3
    Re: 2*pi*i=0?

    in complex analysis, log function is a Multi-valued function,ln(z) = |z| + arg(z), you should choose a single value branch for log, or use Riemann surface
     
  5. Jan 9, 2009 #4

    arildno

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    Re: 2*pi*i=0?

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

    That inference, from line 1 to line 2 is invalid.
     
  6. Jan 9, 2009 #5

    HallsofIvy

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    Re: 2*pi*i=0?

    In "polar form", [itex]z=re^{i \theta}[/itex], [itex]\theta[/itex] represents the angle the line from 0 to z makes with the positive real axis. In that sense, yes, [itex]2\pi[/itex] is the same angle as 0. That's why, as arildno suggested, in complex numbers, the exponential function is no longer single valued. f(x)= f(y) implies x= y only if f is single valued.
     
  7. Jan 9, 2009 #6
    Re: 2*pi*i=0?

    Ok I get it; so it's because complex exponentials have more than one solution.
     
  8. Jan 9, 2009 #7
    Re: 2*pi*i=0?

    I think you mean it's not one-to-one! It's the logarithm that's not single-valued. :)
     
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