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B Can some explain why e^i(x) = cos(x) + isin(x)?

  1. Apr 6, 2016 #1
    Can anyone provide me with a simple explanation for why e^i(x) acts like this.
     
  2. jcsd
  3. Apr 6, 2016 #2
    If you draw a unit circle in the complex plane, the exponential maps the angle theta to the complex x and y coordinates. Keep in mind that by definition, x = cos(theta) and y = sin(theta).

    e^(i theta) = x + iy

    You can also see this when you taylor expand the exponential function. You get real and imaginary terms, and these are your cosine and sine respectively.

    This picture demonstrates the concept better than words imo.
    https://upload.wikimedia.org/wikipe...'s_formula.svg/2000px-Euler's_formula.svg.png
     
  4. Apr 6, 2016 #3

    Ssnow

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    If you try using the Taylor expansion of ##e^{ix}## you will find that this is equal to the Taylor expansion of ##\cos{x}## plus ##i## times the Taylor expansion of ##\sin{x}## ...
    You can find the proof on the major part of Analysis books. This formula is useful in the representation of a complex number ##z##. The remarkable relation ##e^{i\pi}=-1## can be obtained setting ##x=\pi##.
     
  5. Apr 6, 2016 #4
    another remarkable relation is that i^i ~ 0.2. :wink:
     
  6. Apr 6, 2016 #5
    There's no need to be approximate: [tex]i^i=e^{\text{-}\frac{\pi}{2}}[/tex]
     
  7. Apr 6, 2016 #6

    FactChecker

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    Another intuitive explanation: Remember that d/dx( eix ) = i*eix. That means that as real values of x increase from 0, eix starts at ei*0 = 1 and it always moves at an exact right angle to it's current radius vector from 0. So it goes around the unit circle in the complex plane. Looking at its real and imaginary part, you see that they match the cos(x) and i*sin(x), respectively.
     
  8. Apr 6, 2016 #7
    So e^x=1+x+x^2/2!+x^3/3!....continues to infinity
    Replacing x with ix 1+ix-x^2/2! - ix^3/3!...continues to infinity
    separating the real terms from the imaginary:
    1-x^2/2!+x^4/4! - x^6/6!
    factor out i in the imaginary terms
    i(x-x^3/3!+x^5/5!-x^7/7!)
    The real terms match up exactly with the macclaurin expansion of cos(x) whereas the imaginary terms match up with the macclaurin expansion of sin(x) so
    e^ix= cos(x) + i(sin(x))
     
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