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Euler's identity 
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#1
Aug1308, 02:44 AM

P: 195

How does y = e^(x(1i)) + 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) 


#2
Aug1308, 03:21 AM

P: 35

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


#3
Aug1308, 03:27 AM

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#4
Aug1308, 08:07 AM

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Euler's identity



#5
Aug1308, 08:39 AM

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