e^(pi i)

= -1.


Just thought I'd state that.
Cause it just blew my mind.

e^(pi i) = x
ln (e^(pi i)) = ln x
(pi i)(lne) = ln x
pi i = ln x

domain of: x > 0

I must've missed something...

http://en.wikipedia.org/wiki/Euler_identity

e^(pi i) = x
ln (e^(pi i)) = ln x
(pi i)(lne) = ln x
pi i = ln x

domain of: x > 0

I must've missed something...

The "domain" of a real function refers to which real x-values produce real y-values. pi i isn't a real value, so naturally it isn't in the real domain of ln.

The equation follows from the identity e^(ix) = cos(x) + i sin(x), for which proofs are all over the Internet. The function's called cis(x) and has some interesting properties. From its definition it's pretty apparent that cis(pi) = -1, among other things.

Reference: http://mathworld.wolfram.com/EulerFormula.html

Xkcd

Funny thing you mention it, actually. Me and my friend have a habit of taking complicated mathematical formulas and telling them repeatedly to many people in order to annoy them. Over the years we have passed many mathematical subjects, because the guys we were trying to annoy actually learned these subjects in math class. We have already gone through trigonometry, calculus, infinity maths, complex numbers and just a few weeks ago we started using complex numbers with euler's formula (e^ai = cos(a) + isin(a)). Allow me to explain:

e^pi*i = cos(pi) + i*sin(pi) = -1 + 0
e^pi*i + 1 = 0 (according to wikipedia, this formula was voted the most beautiful mathematical formula ever for using all 5 of math's most important constants)
e^i*(pi/2) = cos(pi/2) + i*sin(pi/2) = i
therefore:
i^i = e^i*(pi/2)^i = e^(i^2)*(pi/2) = e^-pi/2
i'th root of i = i^1/i = e^i*(pi/2)^(1/i) = e^(pi/2)
(-1)^-i = i^2^-i = (e^i*(pi/2))^2^-i = e^i*pi^-i = e^-(i^2)*pi = e^pi (also known as gelfond's constant)

We usually tell the above paragraph at least two times a day to random students. Yes, I am quite the annoying type of a person.

*eye twitch*

holy moley that;s a lot of numbers and letters. i understood you up to the euler's formula and then i got lost.

some power formulae:
a^i = (e^ln(a))^i = e^ln(a)i = cos(ln(a)) + isin(ln(a))
also, another thing - the mathematical rule that says (ab)^c = (a^c)(b^c) doesn't work with i. If it did, this is what would happen:
1^i = (-i^2)^i = (-1*i^2)^i = ((-1)^i)(i^2i) = (i^2i)(i^2i) = i^4i = (e^ipi/2)^4i = e^-2pi
this causes major problems with other mathematical rules, as well as the one above which states that:
1^i = cos(ln(1)) + isin(ln(1)) = cos(0) +isin(0) = 1

That's one beautiful mathematical equation.