Let T denote the circle group, that is, the multiplicative group of all complex numbers with absolute value 1. Let f : T → T be a (sequentially) continuous map, and such that f(z2 ) = f2 (z) for all z ∈ T. Then there is an integer k such that f(z) = zk for all z ∈ T.
From The Weierstrass factorization theorem, we know that f have ever any root on T. Also one can use the power series and obtains unknown coefficients with comparing them, similar with attaching picture.
Another proof
let f(z)=exp(ig(a)), such that z=exp(i a), a belongs to [0,2pi]. From
f(z^2)=f^2(z), one concludes that there is an analytical function g(a) having the property as 2g(a)=g(2a)+2npi, for some integer number n, so n is zero. We must answer to this question which g(a) is only either linear function or not!
By power series, we could imply that g is linear function, then g(a)=k.a. indeed
f(z)=z^k, for any positive real number, are alone class functions, they satisfy at this property.
- Badges
- Science topic
- Similar topics
- Mathematics