Use Euler's identity: eiπ+1 = 0

We have, eiπ=-1, then (eiπ)2k=(-1)2k = 1, Q.E.D.

P.S. To find more on Euler's identity please follow the thread.

https://www.researchgate.net/post/What_is_the_most_beautiful_theorem_in_mathematics?_tpcectx=profile_highlights

e(j*thetha)=cos(theta)+j*sin(theta), then e(-j*2pi*k)=cos(-2*pi*k)+j*sin(-2*pi*k) = 1+J*0=1

Helpful.... how about x(-j*thetha), that got negative before j thetha

Dear Chong,

Recall that  cos(2πk) = 1  and  sin(2πk) = 0  for every integer, k. Then,

ei2πk= cos(2πk) + i sin(2πk) = 1, for every integer, k.

Regards.

Use Euler's identity: eiπ+1 = 0

We have, eiπ=-1, then (eiπ)2k=(-1)2k = 1, Q.E.D.

P.S. To find more on Euler's identity please follow the thread.

https://www.researchgate.net/post/What_is_the_most_beautiful_theorem_in_mathematics?_tpcectx=profile_highlights