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I Intuition for Euler's identity

  1. Nov 1, 2016 #1
    I read an intuitive approach on this website. You should read it, it's worth it:

    I read that an imaginary exponent continuously rotates us perpendicularly, therefore, a circle is traced and we end up on -1 after rotating through pi radians.
    If that's true, then why doesn't e^-pi rotates us continuously through 180 degrees so that we end up on the negative axis? '-' has more rotating power than 'i',right?
    And, if that's not true, then please share your intuition of the formula.
  2. jcsd
  3. Nov 1, 2016 #2


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    No. The real part of the exponent does no rotation at all. Suppose we separate the exponent x+iy, into its real part, x, and its imaginary part, iy. Then ex+iy = exeiy, where the factor ex is the usual real exponential and eiy is a pure rotation in the complex plane. ex is just a pure multiplier with no rotation of the vector eiy in the complex plane. So e does no rotation at all.
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