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unreality
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Okay the answer is as follows:

2 pi needs to be expressed as 0 to create an equivalent expression due to the fact that you are working in polar coordinates with imaginary numbers.

You would end up with 0*i = 0 which is true.

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actually - as I understand it ( :P ) - nobody has gotten it yet ;D

I'm pretty sure Scrappy is right, though his answer is a bit hard to follow. In polar coordinates, an angle of 2*pi radians is the same as 0 radians, and the 2*pi must be reduced to 0 to simplify the expression, just as 4*pi and 6*pi, and any 2*n*pi, where n is an integer...no matter how many times you go around the circle, you still end up at 0. When you enter ln(e^(2*n*pi*i)) for any n into a calculator, it gives you an answer of 0 because it automatically reduces the expression.

edit: btw, welcome to the Den, Scrappy!

Edited by lazboy
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yes, so the trig functions reduce it on their own - my calculator doesn't do what you're saying :) So there is at least two fallacies then :o

the problem occurs when you square it. When you square something, the information is lost. Ie:

5^2 = 25 = -5^2

for that reason, when you take the square root of 25, you're not sure whether it's positive or negative 5, so you write ±5

well in this case, when you square -1, if you leave it as (-1)2 instead of +1, then it works:

e = -1

square

e2iπ = (-1)2

take ln of both sides

2iπ = ln((-1)2) = 2*ln(-1)

2iπ = 2*ln(-1)

drop the 2s

iπ = ln(-1)

Which is a true statement of course

edit-typo

Edited by unreality
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You actually have a fallacy in the very first statement. I don't know how to do the funky formatting though, so I'm going to use ^ to represent exponents.

Instead of e^(in) = -1, you should be using e^i(2n+1). After all, if n is an integer, then n can be any integer, including the even ones. The way to represent odd integers isn't n, it's 2n+1. This would modify the proof from there on out, but I can't do the formatting, and don't want to do the math, so someone else can ;)

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You actually have a fallacy in the very first statement. I don't know how to do the funky formatting though, so I'm going to use ^ to represent exponents.

Instead of e^(in) = -1, you should be using e^i(2n+1). After all, if n is an integer, then n can be any integer, including the even ones. The way to represent odd integers isn't n, it's 2n+1. This would modify the proof from there on out, but I can't do the formatting, and don't want to do the math, so someone else can ;)

As unreality said, the n isn't an n its a π (or pi). If you change the font, then its a bit more legible: π π π π

e^(iπ) = -1 (or e^(iπ) + 1 = 0) is a famous equation*, and, for me at least, is a pretty surprising one linking so many fundamental numbers.

I first read the equation as e ^ ln (or natural log) and thought unreality had made a different blunder, meaning to put ln(e) = 1.

*eulers equation, i think

Edited by armcie
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it's called Euler's Identity, and is a special case of Euler's Equation (I guess it's actually called Euler's Formula not Euler's Equation, whatever :P)

To me, it's just as surprising as a formula for i^x I discovered by accident a few months ago:

i^x = cos(x * pi/2) + i*sin(x * pi/2)

that's off the top of my head btw, but I'm pretty sure it's correct. Of course, the special cases give i^2 = -1, i^3 = -i, i^4 = 1, etc, likewise i^(1/2), the square root of i, gives sqrt(2)/2 + i*sqrt(2)/2

edit: more surprising is that i^i is a real number lol

Edited by unreality
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