The Brahmagupta-Fibonacci identity states that for all $a,b,c,d$, $$(a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2.$$ This is easy to prove algebraically, and says essentially that the norm on complex numbers is multiplicative. Is there a geometric proof though? Bak and Newman's textbook "Complex Analysis" suggests there is (p.19), but I can't see how it would go.
Asked
Active
Viewed 801 times
10
-
5This one looks like it comes from Proofs Without Words, but it's not a very good "geometric" proof, since it's basically just algebraic manipulation in geometric clothes. – Chappers Jun 12 '17 at 11:19
-
Actually this is a possible duplicate of https://math.stackexchange.com/questions/1629824/geometric-interpretation-of-complex-algebraic-proof-of-sum-of-squares-statement?rq=1 – Blunka Jun 13 '17 at 00:29
