## Euler's identity

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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 why don't you write your equations using [tex] tabs... it will be more readable by others...

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 Quote by jaejoon89 How does y = e^(x(1-i)) + e^(x(1+i)) work out to y = (e^x)sinx + (e^x)cosx???
It doesn't.

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## Euler's identity

 Quote by jaejoon89 How does y = e^(x(1-i)) + e^(x(1+i)) work out to y = (e^x)sinx + (e^x)cosx???
Obviously, by Euler's identity, (e^x)sinx+ (e^x)cosx= e^(x(1+i)) only, not that sum.

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 Quote by HallsofIvy Obviously, by Euler's identity, (e^x)sinx+ (e^x)cosx= e^(x(1+i)) only, not that sum.
Halls meant (e^x)*i*sinx+ (e^x)cosx. The expansion you did, 2cos(x)e^x, is correct. and it's not equal to sin(x)e^x+cos(x)e^x.