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Smash_Returns
03-11-2009, 09:37 PM
(e=the base of the natural log; i = sqrt(-1))

e^(pi*i) = -1 (true statement)
(square both sides)
e^(2*pi*i) = 1
(take the natural log of both sides)
ln e^(2*pi*i) = ln 1
(simplify the natural log)
2*pi*i*ln e = ln 1
(further simplify; ln e = 1, ln 1 = 0)
2*pi*i = 0
Q.E.D.

Where is the error in the above proof?

sKratch
03-11-2009, 10:10 PM
The argument of the natural log is some crap to a power. The log is not raised to a power. Therefore you can't bring down the exponent.

Smash_Returns
03-11-2009, 10:19 PM
a log brings down a power. e is raised to 2*pi*i. The natural log allows us to bring down the power. That is the whole POINT of a logarithm.

That is not a problem.

Endymion
03-12-2009, 12:28 AM
a log brings down a power. e is raised to 2*pi*i. The natural log allows us to bring down the power. That is the whole POINT of a logarithm.

That is not a problem.

[edit for easier answer]:
you missed the branch cut.

Budzy
03-12-2009, 12:30 AM
you can only bring down a power if it's real. complex powers don't come down like that.

l2math.

My complex powers come down hard on anyone who displeases me.

sKratch
03-12-2009, 07:24 AM
Endy's right; logs and exponentials are significantly different in the complex plane.

Virtuoso
03-12-2009, 02:14 PM
Here's a short (and easy) one:

Assume that a is a positive number.

a - 1 < a | *(-a)
-a^2 + a < -a^2 | +a^2
a < 0

Haha, so lame...

Smash_Returns
03-13-2009, 09:39 AM
Endymion is correct. When we bring down the 2*pi*i, because the power is complex, the modular base changes.