Sunday, April 17, 2011

Expressions with e, i and π

We all know that ... e^(i*π) + 1 = 0


I found out that ....


e^(i*π/3) + e^(2i*π/3) + e^(3i*π/3) + e^(4i*π/3) + e^(5i*π/3) + e^(6i*π/3) = 0



And

e^(i*π/3) + 2e^(2i*π/3) + 2e^(3i*π/3) + 2e^(4i*π/3) + e^(5i*π/3) + 9e^(6i*π/3) = 6

e^(2i*π/3) + e^(4i*π/3) + e^(6i*π/3) = 0

6e^(6i*π/3) = 6

6*[cos(2π) + i*sin(2π)] = 6


6/π^2

It's also the probability that a number picked at random from the set of integers will have no repeated prime divisors



e^(2i*π) = e^(4*i*π) = e^(6*i*π) = 1

e^(3i*π) = -1 .... e^(5i*π) = -1 .... e^(7i*π) = -1


e^(2!*i*π) = e^(3!*i*π) = e^(4!*i*π) = e^(5!*i*π) = 1

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