Identity for exponential function with imaginary arguments

[ParityCheck]

I have seen the following identity used.
Exp[iw/2]-Exp[iw/2]=Exp[iw]-1
I can't find this in any book and I can't prove it myself.
The left side equals 2isin(w/2)
The right side equals cos(w)+isin(w)-1
On the face of it, that seems to make the identity absurd
How can one go about proving such an identity? Or is it wrong?

 

[tiny-tim]

Welcome to PF!

Exp[iw/2]-Exp[iw/2]=Exp[iw]-1
…
How can one go about proving such an identity? Or is it wrong?

Hi ParityCheck ! Welcome to PF!

(e^iw/2 - e^-iw/2)e^iw/2 = e^iw - 1

 

[ParityCheck]

Thanks. I was making it way too complicated with trig identities and series and...