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Math Help - e^ipi squared

  1. #1
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    e^ipi squared

    today in class i was asked what was wrong with this
    e^ipi + 1=0
    e^ipi=-1
    e^2ipi=1
    2ipi=ln1
    this is obviously wrong, is it the stage where you square e^ipi,
    i havnt come across complex powers before. thanks for any help sorry if this is a stupid question.
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  2. #2
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by hmmmm View Post
    today in class i was asked what was wrong with this
    e^ipi + 1=0
    e^ipi=-1
    e^2ipi=1
    2ipi=ln1
    this is obviously wrong, is it the stage where you square e^ipi,
    i havnt come across complex powers before. thanks for any help sorry if this is a stupid question.

    you mean?

    e^{\pi} + 1 = 0

    if you mean this, then it's obviously wrong because it isn't 0 xD

    e^{\pi} + 1 = 24.14069263...
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by hmmmm View Post
    today in class i was asked what was wrong with this
    e^ipi + 1=0
    e^ipi=-1
    e^2ipi=1
    2ipi=ln1
    this is obviously wrong, is it the stage where you square e^ipi,
    i havnt come across complex powers before. thanks for any help sorry if this is a stupid question.
    Hello hmmmm,

    See here: Euler's identity - Wikipedia, the free encyclopedia
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  4. #4
    Moo
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    No, e^{i \pi}+1=0 and that's true

    The problem is when you take the logarithm.
    Because e^{i \alpha}=e^{i (\alpha+2k \pi)} where k is an integer, this is why your step is wrong.

    So following this reasoning, you can have e^{2i \pi}=e^{i(2 \pi-2\pi)}=e^0=1 which is indeed true.

    You can have a look at this : Complex logarithm - Wikipedia, the free encyclopedia
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  5. #5
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    thanks

    thanks but if possible could you elaborate on your third line?
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  6. #6
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    Quote Originally Posted by hmmmm View Post
    thanks but if possible could you elaborate on your third line?
    Imagine that e^{i \alpha}=\cos(\alpha)+i \sin(\alpha) (this is a definition)
    Since the cosine and the sine are 2 \pi-periodic functions, we can write :
    e^{i \alpha}=\cos(\alpha)+i \sin(\alpha)=\cos (\alpha+2k \pi)+i \sin(\alpha+2k \pi)=e^{i (\alpha+2k \pi)}
    The logarithm in the complex values has an infinity of values, unless you define a continuous logarithm (but I don't think this is something you need to know yet).

    another way of seeing it is that 1=1+i 0=\cos(2 k \pi)+i \sin (2 k \pi)=e^{i 2k \pi}
    So you can multiply e^{i \alpha} by e^{i 2 k \pi}, which gives e^{i (\alpha+2k \pi)}, it won't change the value.
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  7. #7
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    Cool

    Hello, hmmmm!

    Today in class i was asked what was wrong with this:

    . . e^{i\pi} + 1\:=\:0

    . . e^{i\pi} \:=\:-1

    . . e^{2i\pi}\:=\:1 . . . square both sides

    . . 2i\pi\:=\:\ln1 \:=\:0

    This is obviously wrong.
    Is it the stage where you square e^{i\pi} ? . . . . yes

    We have: . e^{i\pi} \:=\:-1

    Square: . e^{2i\pi} \:=\:(-1)^2

    Then: . 2i\pi \:-=\:\ln(-1)^2 \:=\:2\ln(-1)

    Hence: . i\pi \:=\:\ln(-1) . . . and this is true.

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  8. #8
    Moo
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    Hence: . i\pi \:=\:\ln(-1) . . . and this is true.
    I'm sorry to say, but it's not always true. It depends on the logarithm you take.

    You can define a logarithm such that \ln(-1)=3i \pi
    Read my explanation (or what it tries to be) above.
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