Proof that e^i(pi) = -1
Date: 06/02/99 at 00:48:10 From: Nathan White Subject: e^[i(pi)] = -1 How can it be proven that e^[i(pi)] = -1? And why does it matter?
Date: 06/02/99 at 06:28:07 From: Doctor Mitteldorf Subject: Re: e^[i(pi)] = -1 Dear Nathan, When you do algebraic manipulations, you want to be sure that everything that you do legally "makes sense," which means that equations should have solutions, so you're not just casting around in vain. You can easily write real equations that don't have real solutions. But if you expand to complex numbers, then all algebraic equations become solvable. How then do you make sense out of complex numbers in the exponent? The Rosetta stone is the Euler equation: e^(ix) = cos(x) + i sin(x) You can verify that this equation makes a sensible definition by expanding the two sides as Taylor series in x. You can also differentiate both sides and see that the answer is self-consistent. Thirdly, you can use the formula for cos(2x) and sin(2x) to show that the right side has the property you expect from an exponential, so that e^[i(2x)] = [e^(ix)]^2. Once you believe the Euler equation, you have a pretty easy time of it. If you let x = pi/2, then the real part is zero e^[i(pi)] = cos(pi) + i sin(pi) = -1 + 0*i If you are happy to accept integration of complex functions, a quick and reasonably rigorous proof follows (from Dr. Anthony's answer in the Dr. Math archives): Euler Equation http://mathforum.org/dr.math/problems/dale9.13.97.html Let z = cos(x) + i.sin(x) dz/dx = -sin(x) + i.cos(x) = i[cos(x) + i.sin(x)] dz/dx = i.z so dz/z = i.dx INT[dz/z] = INT[i.dx] ln(z) = i.x + const z = e^(i.x + const) Now from z = cos(x) + i.sin(x) when x=0, z=1 so the constant = 0 Thus z = e^(i.x) That is cos(x) + i.sin(x) = e^(i.x) When x = pi -1 = e^(i.pi) 0 = e^(i.pi) + 1 or e^(i.pi) + 1 = 0 This equation combines the 'famous five' most important numbers in mathematics, 0, 1, e, i, pi. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.