Euler's identity


by jaejoon89
Tags: euler, identity
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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xavier_r
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#2
Aug13-08, 03:21 AM
P: 35
why don't you write your equations using [tex] tabs...
it will be more readable by others...
Defennder
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#3
Aug13-08, 03:27 AM
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Quote Quote by jaejoon89 View Post
How does y = e^(x(1-i)) + e^(x(1+i))
work out to y = (e^x)sinx + (e^x)cosx???
It doesn't.

HallsofIvy
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#4
Aug13-08, 08:07 AM
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Euler's identity


Quote Quote by jaejoon89 View Post
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.
Dick
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Aug13-08, 08:39 AM
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Quote Quote by HallsofIvy View Post
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.


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