How Does Euler's Identity Simplify the Expression y = e^(x(1-i)) + e^(x(1+i))?

[jaejoon89]

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)

 

[xavier_r]

why don't you write your equations using $$tabs...<br /> it will be more readable by others...$$

 

[Defennder]

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]

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]

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.