Ronald Connelly Blog
Sunday, July 5, 2015
Complex exponential product
Show that, for all real p and m,
e^{2mi\cot^{-1}(p)}\left(\dfrac{pi+1}{pi-1}\right)^m=1
No comments:
Post a Comment
Newer Post
Older Post
Home
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment