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How can i solve this KIND OF VALUE (〖-i)〗^i
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we might write -i as cos 3pi/2 + i sin 3pi/2 = e^(i 3pi/2) so -i^i = [e^(i 3pi/2)]^i = e^ -3pi/2 = approx 0.00898
a rational number Surprising result.
is that imaginary number?
well i is but 00898 isn't
oh ok, i thought it was just an imaginary number to the power of i is,,that will be equal to 1.
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I should have said 0.00898 is a real number - it might not be rational.
there are some remarkable results sometimes when you play with complex numbers. the Euler equation is derived from comples numbers also e^i pi = cos pi + i sin pi = -1 + 0i = -1 so e ^ i pi + 1 = 0 This equation unites the 5 most important numbers in mathematics: 0 , 1, pi, e, and i.
Thanks DEAR
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