I don't understand this question at all. I've asked it a few times, and have gotten what look like a correct solutions, but I don't understand the reasoning behind many of the steps. I'm asking it again, hoping to get an even more thorough explanation. I'm sorry, it just isn't making sense to me. Thanks for your time. 9. Let be a bounded open subset of C, and 4:Prove that if there exists a point zo EN such that2 a holomorphic function.4(20) = 20 and ' (zo) = 1then is linear.[Hint: Why can one assume that zo = 0? Write y(z)=z+anz" +O(z+¹) near0, and prove that if yk = o...o (where y appears k times), then k(z) =z+kanzn +0(z+¹). Apply the Cauchy inequalities and let k → ∞ to concludethe proof. Here we use the standard O notation, where f(z) = O(g(z)) as z → 0means that |ƒ(z)| ≤ C|g(z)| for some constant C as [z] → 0.]

9. Let 2 be a bounded open subset of C, and y: → a holomorphic function. Prove that if there exists a point zo E such that 4(20) <= 20 and (20) = 1 then 4 [Hint: Why can one assume that zo = 0? Write y(z)=z+anz" +O(z+¹) near 0, and prove that if k = 90.0 (where appears k times), then k(z) = z+kanz" +0(z+1). Apply the Cauchy inequalities and let k→ ∞ to conclude the proof. Here we use the standard O notation, where f(z) = O(g(z)) as z → 0 means that |ƒ(z)| ≤ C|g(z)| for some constant C as [z] → 0.] is linear.

9. Let 2 be a bounded open subset of C, and y: → a holomorphic function. Prove that if there exists a point zo E such that 4(20) <= 20 and (20) = 1 then 4 [Hint: Why can one assume that zo = 0? Write y(z)=z+anz" +O(z+¹) near 0, and prove that if k = 90.0 (where appears k times), then k(z) = z+kanz" +0(z+1). Apply the Cauchy inequalities and let k→ ∞ to conclude the proof. Here we use the standard O notation, where f(z) = O(g(z)) as z → 0 means that |ƒ(z)| ≤ C|g(z)| for some constant C as [z] → 0.] is linear.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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