EDIT I don’t know why I thought computing a series expansion for using Euler’s identity was clever, while it’s just way simpler via the usual expansion of in the real numbers… I’m a bit ashamed of this post.
The classic identity works for all complex values of . Hence, by setting we find Therefore, we have Distributing yields Notice that in the previous sum, each pair where is odd and is even cancels out the term corresponding to the pair . Hence we may sum over and that have the same parity: