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Euler's identity
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jaejoon89
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1
Aug13-08, 02:44 AM
P: 195
How does y = e^(x(1-i)) + e^(x(1+i))
work out to y = (e^x)sinx + (e^x)cosx???
Using Euler's identity I get,
y = (e^x)e^-ix + (e^x)e^ix
y = e^x(cosx - isinx + cosx + isinx)
y = e^x(2cosx)
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