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0=i2pie ?

  1. May 20, 2012 #1
    0=i2pie !!!???

    We know that any number to the power of zero equals one. Ex; e^0=1

    But we also know that eulers number (e=2.718...) to the power of i (√-1) times 2∏ equals one; e^i2∏=1

    So we have to equations that both equals 1. That means that e^0=e^i2∏ and that;

    0=i2∏

    Is this right? (it can't be?)
     
  2. jcsd
  3. May 20, 2012 #2
    Re: 0=i2pie !!!???


    Take a wild guess...:>)

    What is true is: any non-zero number to the power of zero is one, and [itex]\,e^{2n\pi i}=1\,[/itex] for any integer n...

    DonAntonio
     
  4. May 20, 2012 #3
    Re: 0=i2pie !!!???

    Ok, im in shock!

    If i continue then;

    0=i2pie ⇔ 0/2pie=i ⇔ 0=i !!

    If this is true, doesn't it imply that i can f.ex write; 0=0+ai where a is any real number?

    and f. ex 3=3+4i. (or 3=3+ai where a is any real number)

    this must be true if 0=i2pie is true.

    (I think?)

    feel free to disagree, honestly;)
     
    Last edited: May 20, 2012
  5. May 20, 2012 #4
    Re: 0=i2pie !!!???

    and that one can write b=ai where a and be are two real numbers that doesn't necessary equal!

    f.ex 3=298i

    ?????????
     
  6. May 20, 2012 #5
    Re: 0=i2pie !!!???

    You are working under the assumption that ea = eb implies that a = b for all complex numbers a and b. Can you prove this assertion? If not, can you prove that it is false?
    Consider the fact that cos(2Pi) = cos(0). Does this imply that 2Pi = 0? If not, why not? Is this property applicable to your case?
     
  7. May 20, 2012 #6

    arildno

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    Re: 0=i2pie !!!???

    Is -1=1 just because (-1)^2=1^2???
     
  8. May 20, 2012 #7
    Re: 0=i2pie !!!???

    That wasn't how i taught of it;

    If one can say that 0=i it must be because when "you are" at any point ON the imaginary axis, then the value on the real axis will always be zero.

    Just as when you say the same in a typical graph with x- and y-axis. When "you are" at any point on the y-axis, then x=0.
     
  9. May 20, 2012 #8

    dextercioby

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    Re: 0=i2pie !!!???

    You need to revise your complex numbers . e^z (with z complex) is not injective.
     
  10. May 20, 2012 #9
    Re: 0=i2pie !!!???

    Translating for the OP, who may not have any knowledge of set theory: Given two complex numbers w and z, ez = ew does not imply that z = w.
     
  11. May 20, 2012 #10
    Re: 0=i2pie !!!???

    The complex exponential is periodic function with period 2pi. So when two complex exponential are equal, that means the exponents are equal up to an additive constant of form n2pi, where n is integer.
    In mathematical terms:
    e^ix = e^iy => x = y + n2pi
     
  12. May 20, 2012 #11
    Re: 0=i2pie !!!???



    Well, in fact it is [tex]e^{z_1}=e^{z_2}\Longleftrightarrow z_1=z_2+2n\pi i\,\,,\,\,z_1,z_2\in\mathbb{C}[/tex] no need to add that "i" in th exponent

    DonAntonio
     
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