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Homework Statement
Suppose that f is continuous and that there exist constants A,B ≥ 0 and k>1 such that |f(z)|≤A|z|−k for all z such that |z|>B. let CR denote the semicircle given by |z| = R, Re(z) ≥ 0. Prove that limR→∞∫f(z)dz=0
It is probably in your textbook, but just not named. Many texts (c.f. Mathematical Methods for Physicists A Concise Introduction Tai L. Chow p. 255) simply list it as a property of complex integrals and give a short proof of it.I havent done the estimation lemma but I looked it up
Let f : U[itex]\rightarrow[/itex]C be continuous (where U is some subset of C), let [itex]\gamma[/itex] be a path in U, and suppose |f(z)| < M for all z [itex]\in \lambda[/itex]. Let length( [itex]\lambda[/itex])= L. Then
|[itex]\int[/itex]f(z)dz| [itex]\leq[/itex] ML.
Right or, using the estimation lemma with the fact that [itex]|f(z)| \leq A|R|^{-k} [/itex], you haveThe arc length is[itex]\pi r[/itex]
[itex] |f(z)| \leq A|z|^{-k} [/itex] the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of [itex] \frac{A}{|z|^{k}} [/itex] gets smaller so as R[itex]\rightarrow\infty[/itex] then [itex] |f(z)| \rightarrow [/itex] 0