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Is 5pi/2=pi/2 ?

  1. Jun 15, 2010 #1
    Hi,

    [tex]i=e^{\frac{\pi }{2}i}=e^{\frac{5\pi }{2}i}\Rightarrow e^{\frac{\pi }{2}ii}=e^{\frac{5\pi }{2}ii}=e^{-\frac{\pi }{2}}=e^{-\frac{5\pi }{2}}[/tex]

    The result of this is 5pi/2=pi/2.Please explain to me.
     
  2. jcsd
  3. Jun 15, 2010 #2
    You need to be careful with the function a -> a^i since you need logarithm to define it and therefore need to stick with one branch.
     
  4. Jun 16, 2010 #3

    Landau

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    exp(a)=exp(b) does not imply a=b, but a=b+i2pi.
     
  5. Jun 16, 2010 #4
    cos(x)=cos(-x).
    That is a true statement for all real values of x.
    So x = -x for any real value of x, which implies that every real number is actually EQUAL TO ZERO!!!

    Do you (OP) understand the error in this??
     
  6. Jun 16, 2010 #5
    The problem here is that the exponential function on the complex plane is not injective. That is, even if ea = eb, we cannot infer that a = b. Analogously, even if sin(0) = sin([tex]\pi[/tex]), we cannot infer that 0 = [tex]\pi[/tex].


    Probably a typo there. a and b differ by a multiple of 2 i pi, i.e. a = b + 2k i pi where k is an integer.
     
    Last edited: Jun 16, 2010
  7. Jun 17, 2010 #6

    Mute

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    This is not quite a resolution to the OP's question (even fixed to read [itex]2\pi k[/itex] instead of just [itex]2\pi[/itex]). If a and b are real, that still implies a = b, and k = 0.

    The fully correct resolution is, as eok20 stated, that stating [itex]i = e^{something}[/itex], what you're really doing is writing [itex]i = \exp[\log i][/itex], so you need to choose a branch of the logarithm. In doing so, you cannot equate [itex]\exp(i\pi/2)[/itex] and [itex]\exp(5\pii/2)[/itex] because they exist on different branches.
     
  8. Jun 22, 2010 #7
    The expression of a complex number in exponential form is based on Euler's formula relating it to its polar form: r*exp(i*a)=r*[cos(a)+i*sin(a)]. Because the trig functions are periodic, there is not a unique polar representation. As already stated, there is no injective map.
    Accordingly, the polar form of a complex number is defined based on the principal value of the argument of the complex number, which means restricting a to the interval (-pi,pi].
    So, fundamentally, it is not correct to say that i= exp(i*pi/2) = exp(i*5pi/2).
    However, it so happens that 5pi/2 = pi/2 since 5pi/2= pi/2 + 2pi on the unit circle.
     
  9. Jun 22, 2010 #8
    Is this a serious question?
     
  10. Jun 22, 2010 #9
    Is this a serious answer????
     
  11. Oct 20, 2010 #10
    Hi, I am currently doing Logarithmic Differntiation in my class and had to simplify a similar answer and was able to do so by takeing the natural log of both sides.
    ex: e^x=e^pi/2
    lne^x=lne^pi/2
    x=pi/2
     
  12. Oct 20, 2010 #11
    Be careful about the domain of the functions.
     
  13. Oct 20, 2010 #12
    is 5x/2=x/2 ???
     
  14. Oct 21, 2010 #13

    Mentallic

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    Your point being? Obviously [itex]x=\pi/2[/itex] since the exponents need to be the same. This thread is talking about exponentials in the complex plane, which is a little more complicated.
     
  15. Oct 24, 2010 #14
    Or maybe the question is simple, and the student is just confused by the results. 5pi/2=pi/2 because 5pi=pi because pi is 180 degrees and all the multiples of 2 pi drop out.

    Just a thought.
     
  16. Oct 24, 2010 #15

    Mentallic

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    No [tex]5\pi/2\neq \pi/2[/tex], obviously. What you're referring to is trigonometry such that [tex]sin(\pi/2)=sin(5\pi/2)[/tex] and such. This is not the same.

    I'm still confused as to what GeoFiend was trying to accomplish with his post.
     
  17. Oct 25, 2010 #16

    HallsofIvy

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    Just to stick in my oar: [itex]\pi/2= 5\pi/2[/itex] modulo [itex]2\pi[/itex]. And, if you are working with trig functions that have period [itex]2\pi[/itex], then "modulo [itex]2\pi[/itex]" is sufficient.
     
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