Euler Equation and DeMoivre's Theorem

Date: 05/18/99 at 19:01:03
From: Anthony
Subject: e, pi, and i

Is there a proof of e^(i*Pi) +1 = 0?

Date: 05/19/99 at 09:44:10
From: Doctor Anthony
Subject: Re: e, pi, and i

The derivation is not too difficult if you are familiar with the 
basics of complex numbers and exponential functions.

Start with  z = cos(x) + i*sin(x)    ......(1)

     Then  dz/dx = -sin(x) + i*cos(x)

                 = i(cos(x) + i*sin(x))         (since i^2 = -1)

                 = i*z

       So   dz/z = i*dx


Now integrate both sides

               ln(z) = i*x + c    From (1); when x=0, z=1   so c=0

               ln(z) = i*x

                   z = e^(i*x)    but z = cos(x) + i*sin(x), So

   cos(x) + i*sin(x) = e^(i*x)       ......(2)

Put x = pi in (2) and we get:

        -1 + 0 = e^(i*pi)

and so     e^(i*pi) + 1 = 0

This is the Euler equation.

Now returning to (2) we have

     [cos(x)+i*sin(x)]^n = [e^(i*x)]^n

                         =  e^(i*nx)

                         =  cos(nx) + i*sin(nx)

and this is the statement of DeMoivre's theorem.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/