Q:

Replacing $\sin(z)$ with $1 - e^{2iz}$

I have seen many integral evaluations within logs where they change the sine to:
$$\sin(z) \rightarrow 1 - e^{2iz}$$
Such as here: Contour integral evaluation.
I dont understand how those expressions are equal. Please take a look?
Thanks.

A:

They are not equal. What is true is that $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=\frac{-e^{-iz}}{2i} (1-e^{2iz})$.

