Euler Equation and DeMoivre's TheoremDate: 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/ |
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