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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   

       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   
Associated Topics:
High School Imaginary/Complex Numbers
High School Transcendental Numbers

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