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### Proof that e^i(pi) = -1

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

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