Prove that if two analytic functions in a region G coincide on a subset of G, which has an accumulation point of G, then agree on all points of G.
Thanks for your help
Thanks for your help
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Let F(z) and G(z) be your two functions, and let z0 be your accumulation point, with F(z0) = G(z0). By shifting your functions to f(z) = F(z+z0), and g(z) = G(z+z0), you may assume that z0 = 0.
Let h(z) = f(z) - g(z). Then h(z) vanishes on your subset of G, and h(0) = 0. It suffices to show that h(z) vanishes on all of G.
Since h(z) is analytic write out it's power series: h(z) = ∑ a(n)z^n with n>0. Let a(m) be the first non zero term in this series. Then h(z) = a(m)z^m + z^(m+1)∑ a(n)z^(n-m-1) with n> m. Note that ∑ a(n)z^(n-m-1) converges, so let M = max |∑ a(n)z^(n-m-1)| on your subset.
Select z's in your subset close enough to 0 so that |Mz| < |a(m)|*e where e is arbitrary (this can be done since 0 is an accumulation point of your subset). Then |h(z)| > |a(m)z^m| - |z^(m+1)∑ a(n)z^(n-m-1)| > 0 contradicting that h(z) is uniformly 0 on your subset.
Let h(z) = f(z) - g(z). Then h(z) vanishes on your subset of G, and h(0) = 0. It suffices to show that h(z) vanishes on all of G.
Since h(z) is analytic write out it's power series: h(z) = ∑ a(n)z^n with n>0. Let a(m) be the first non zero term in this series. Then h(z) = a(m)z^m + z^(m+1)∑ a(n)z^(n-m-1) with n> m. Note that ∑ a(n)z^(n-m-1) converges, so let M = max |∑ a(n)z^(n-m-1)| on your subset.
Select z's in your subset close enough to 0 so that |Mz| < |a(m)|*e where e is arbitrary (this can be done since 0 is an accumulation point of your subset). Then |h(z)| > |a(m)z^m| - |z^(m+1)∑ a(n)z^(n-m-1)| > 0 contradicting that h(z) is uniformly 0 on your subset.