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Math Help - Can the ratio of a non-existent number to an imaginary number....

  1. #1
    Senior Member sfspitfire23's Avatar
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    Can the ratio of a non-existent number to an imaginary number....

    ah, thanks guys
    Last edited by sfspitfire23; June 6th 2011 at 07:17 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    e^[(pi)(i)] + 1 = 0
    e^[(pi)(i)] = -1
    (pi)(i) = ln(-1)
    pi = [ln(-1)]/i
    pi = C/d where C = a circle's circumference and d = a circle's diameter

    Thus:
    C/d=[ln(-1)]/i
    If:
    “C” and “d” must be positive real numbers when finding a circle’s area
    ln(-1) = Does not exist
    i = √(-1) = Imaginary number
    Then:
    Can the ratio of a non-existent number to an imaginary number equal the ratio of two real numbers? If not, then can pi still be considered a real transcendental number or is it non-existent and/or imaginary?
    Actually, ln(-1) does exist, just not as a real number. You have calculated it correctly. (Actually ln(-1) = i(2n + 1)*pi where n is any integer, but we won't quibble about that.)

    -Dan
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  3. #3
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    many natural logs of -1 exist. the exponential function isn't 1-1 on the complex plane, so we have multiple choices of domains for a (restricted) inverse.

    if we choose the principal branch, then in fact, \ln(-1) = i\pi.

    see here: Complex logarithm - Wikipedia, the free encyclopedia
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