e^(i*pi) = -1 -- Euler
Proof:
First, we notice,
e^(ix) = cos(x) + i*sin(x)
Let x = pi
e^(i*pi) = cos(pi) + i*sin(pi)
= -1 + 0
= -1
First, we notice,
e^(ix) = cos(x) + i*sin(x)
Let x = pi
e^(i*pi) = cos(pi) + i*sin(pi)
= -1 + 0
= -1
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