- #1
jaejoon89
- 195
- 0
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)
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)