# Residue Theorem

## 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

## The Attempt at a Solution

I dont understand what i'm prooving here. I think it has something to do with jordans lemma but I havent a clue! any ideas...please...

gabbagabbahey
Homework Helper
Gold Member
It is against forum rules to create multiple threads for the same problem.

Have you learned the estimation lemma yet?

Hi!
Sorry about the multiple thread but I called it jordans lemma then thought, it might have nothing to do with it but I couldnt delete the thread after creating it!

I havent done the estimation lemma but I looked it up
Let f : U$\rightarrow$C be continuous (where U is some subset of C), let $\gamma$ be a path in U, and suppose |f(z)| < M for all z $\in \lambda$. Let length( $\lambda$)= L. Then

|$\int$f(z)dz| $\leq$ ML.

vela
Staff Emeritus
Homework Helper
I deleted the other thread. Next time, just hit the report button and ask a mentor to edit or delete the thread for you.

gabbagabbahey
Homework Helper
Gold Member
I havent done the estimation lemma but I looked it up
Let f : U$\rightarrow$C be continuous (where U is some subset of C), let $\gamma$ be a path in U, and suppose |f(z)| < M for all z $\in \lambda$. Let length( $\lambda$)= L. Then

|$\int$f(z)dz| $\leq$ ML.

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.

What is the arclength $L$ of your semicircular contour CR? If $|f(z)| \leq A|z|^{-k}$ (which is what I assume you meant when you wrote |f(z)|≤A|z|−k) and $|z|>B>0$, what is an upper bound for $|f(z)|$ on your semi-circular contour?

What does that tell you about $\left| \int_{CR} f(z) dz \right|$?

The arc lenght is$\pi r$

$|f(z)| \leq A|z|^{-k}$ the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of $\frac{A}{|z|^{k}}$ gets smaller so as R$\rightarrow\infty$ then $|f(z)| \rightarrow$ 0
I guess thats what ther're saying but i guess i better try and word it better

gabbagabbahey
Homework Helper
Gold Member
The arc length is$\pi r$

$|f(z)| \leq A|z|^{-k}$ the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of $\frac{A}{|z|^{k}}$ gets smaller so as R$\rightarrow\infty$ then $|f(z)| \rightarrow$ 0

Right or, using the estimation lemma with the fact that $|f(z)| \leq A|R|^{-k}$, you have

$$\left| \lim_{R \to \infty} \int_{CR} f(z)dz \right| = \lim_{R \to \infty} \left| \int_{CR} f(z)dz \right| \leq \lim_{R \to \infty} \left( A|R|^{-k} \right) = 0$$