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Physical explanation of e^i*pi=-1

  1. Jan 12, 2015 #1
    Hi everyone ,
    i was wondering for a while to get a satisfactory proof of the equation : e^i*pi = -1
    yes , i know it can be derived from euler's formula ...
    which is e^i*x = cosx + isinx ( which can be proved using differential calculus )
    so, e^i*pi = cos(pi) + isin(pi) ( which leads to the result we get )
    but , the problem is ,i can't seem to get the physical explanation of this equation ( e.g. what e^i*pi = -1 exactly means in reality ) and the process we use to obtain it is extremely abstract( at least it seems to me ) .... is there any proof of this equation which makes a bit more sense ?

    my second question is :
    how do we evalute the meaning of sini and cosi ?
    (suppose if i take x = i then we get ,
    e^i*i = cosi + isini
    =>e^-1 = cosi + isini
    ( if i am not wrong this equation can be solved to get individual values for sini and cosi ... but what does a complex angle mean in the first place in ? ) ... i would appreciate answers with proper physical explanation ....

    thanks ALL,

  2. jcsd
  3. Jan 13, 2015 #2
    I'm not sure of how much of a "physical" explanation it is, but think of if as a vector in the complex plane by letting [itex]i[/itex] be the "imaginary" unit vector [itex]\vec i[/itex] and, say, [itex]\vec r[/itex] the horizontal "real" unit vector. Thus [itex]\vec v = \vec r \cos{x} + \vec i \sin{x}[/itex] represents such a vector in the complex plane. This is what [itex]\text{e}^{i x}[/itex] represents, the "real" unit vector just isn't explicitly written.

    It can be noted that the unit vector [itex]\vec v[/itex] can represent any direction in the complex plane (for a real x, anyways). Changing the value of the argument merely "rotates" the vector around origin. So setting [itex]x = \pi[/itex] gives us the direction represented by the vector [itex]\vec v = - \vec r[/itex].

    As for your second question, [itex]\cos z[/itex] and [itex]\sin z[/itex] for some complex [itex]z = a + i b[/itex] can be evaluated if you use their angle sum identities as well as [itex]\cos x = \frac{\text{e}^{i x} + \text{e}^{- i x}}{2}, \sin x = \frac{\text{e}^{i x} - \text{e}^{- i x}}{2 i}[/itex]. Or you can just directly use [itex]\cos z = \frac{\text{e}^{a}\text{e}^{i b} + \text{e}^{-a}\text{e}^{- i b}}{2}, \sin z = \frac{\text{e}^{a}\text{e}^{i b} - \text{e}^{-a}\text{e}^{- i b}}{2 i}[/itex].
    Last edited: Jan 13, 2015
  4. Jan 13, 2015 #3


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    When you use purely imaginary arguments for trig functions they become hyperbolic functions.

    example cos(ix) = cosh(x).
  5. Jan 14, 2015 #4
    We can think of a series of small rotations starting at 1 and continuing by infinitely small shifts until having traveled the entire length of the curve connecting 1 and -1 in the complex plane. Each shift is expressed as:

    ##1 + i \delta##, where ##\delta## is a small angle.

    The total number of shifts needed, then, is ##\frac{\pi}{\delta}##. Thus, the transformation for the whole rotation is:

    ##(1 + i \delta)^{\frac{\pi}{\delta}}##

    Now, letting ##\delta \to 0## and replacing it by ##\frac{1}{n}## in order to use the definition:

    ##e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n##,

    we arrive at Euler's identity, where ##\pi## is the angle that connects 1 and -1 in the complex plane.
  6. Jan 17, 2015 #5


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    If you start out with eix (x real) and use the standard polynomial expansion for ez (1 + z/1 + z2/2!..., change z into ix and remember that i2= -1, you end up with an expansion containing both real and imaginary coefficients. Separate the real parts into one expansion and the imaginary parts into another. The real part turns out to be the familiar expansion of cos(x) and the imaginary part turns out to be the familiar expansion for sin(x) (with i as a common factor).

    Based on these expansions (which, by the way , converge for all values of x), we have eix = cos(x) + i*sin(x) (for x real). Now pi is real, so ei*pi = cos(pi) + i*sin(pi) = -1 +i*0 = -1.

    The second part of your question: If you change the sign of x in the formula, you get ei*(-x) = cos(-x) + i*sin(-x) = cos(x) - i*sin(x) (basic trigonometric identities).

    Therefore eix + e-ix = 2*cos(x) and eix - e-ix = 2*i*sin(x) which gives you the formula cos(x) = ½(eix + e-ix). Now you want the value of cos(i). Just use the formula: cos(i) = ½(ei*i + e-i*i) =½(e-1 + e1). The formula for sin(x) will give you the value for sin(i).
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