iamqsqsqs Messages 9 Reaction score 0 Thread starter Mar 29, 2012 #1 Suppose f is a biholomorphic mapping from Ω to Ω, if f(a) = a and f'(a) = 1 for some a in Ω, can we prove that f(z) = z for all z in Ω?
Suppose f is a biholomorphic mapping from Ω to Ω, if f(a) = a and f'(a) = 1 for some a in Ω, can we prove that f(z) = z for all z in Ω?
mathwonk Science Advisor Homework Helper Messages 12,042 Reaction score 2,343 Mar 29, 2012 #2 look at the corresponding result for the unit disc, then look at the riemann mapping theorem that says every simply connected proper open set in the plane is equivalent to the disc. (i don't know the answer.)
look at the corresponding result for the unit disc, then look at the riemann mapping theorem that says every simply connected proper open set in the plane is equivalent to the disc. (i don't know the answer.)
micromass Staff Emeritus Science Advisor Homework Helper Insights Author Messages 22,170 Reaction score 3,335 Mar 29, 2012 #3 Schwarz lemma ( http://en.wikipedia.org/wiki/Schwarz_lemma ) can be used to prove that the only biholomorphic mappings from the unit disk to itself have the form [itex]\varphi(z)=\zeta \frac{z-a}{\overline{a}z-1}[/itex] with [itex]|\zeta|=1[/itex] and a in the unit disk. Use this.
Schwarz lemma ( http://en.wikipedia.org/wiki/Schwarz_lemma ) can be used to prove that the only biholomorphic mappings from the unit disk to itself have the form [itex]\varphi(z)=\zeta \frac{z-a}{\overline{a}z-1}[/itex] with [itex]|\zeta|=1[/itex] and a in the unit disk. Use this.