Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The bound $$\left| \int_C f(z)\ dz \right| \le ML$$ where $M$ is the supremum of $\left|f(z)\right|$ on $C$ and $L$ is the length of $C$.

  1. Use (1) to give a bound for the integral of $f(z) = z^n$ ($n$ an integer, possibly negative) on the circle $\left|z\right| = R$.
  2. Use your bound from 1 to describe the behavior of the integrals as $R \rightarrow \infty$ and $R \rightarrow 0$ for different values of $n$. (One value of $n$ stands out. Which?)
  3. Use your comments in response to 2 and Cauchy's Theorem to compute the integrals without parameterizing the circle (except for the exceptional case).

Cauchy's: if $C$ is piecewise smooth and $f$ is differentiable in an open set containing $C$ and the interior of $C$, then $$\int_C f(z)\ dz=0.$$

  1. is easy: the integral $ML = 2 \pi R^{n+1}$.
  2. if $n \ge 0$ and $R \rightarrow \infty$, then $ML= 2 \pi R^{n+1} \rightarrow \infty$, and as $R \rightarrow 0$, then $ML= 2 \pi R^{n+1} \rightarrow 0$. But if $n \le -2$, then as $R \rightarrow \infty$, $ML \rightarrow 0$ and as $R \rightarrow 0$, $ML \rightarrow \infty$. Finally, if $n=-1$, then $n+1=0$ (duh!) and $R^{n+1}=R^0=1$. In that case as, for both $R \rightarrow \infty$ and $R \rightarrow 0$, $ML=2 \pi R^{n+1}=2 \pi R^0=2\pi \rightarrow 2 \pi$.

In summary:$$ \begin{array}{|c||c|c|} \hline \lim(2 \pi R^{n+1}) & \text{as } R \rightarrow \infty & \text{as } R \rightarrow 0 \\ \hline n \leq -2 & 0 & \infty \\ n=-1 & 2 \pi & 2 \pi \\ n \geq 0 & \infty & 0 \\ \hline \end{array}$$ 3. How does Cauchy's Thm apply to this?

share|cite|improve this question
Maybe you should use this variant of Cauchy's Theorem: – Beni Bogosel Apr 5 '12 at 11:07
If I recall there is a very trivial solution to this but I don't want to give the game away. Suffice to say that there is a strong result that arises for $C$ a closed curve. Perhaps rewrite $z$ in complex exponential form. Does that help you? – Autolatry Apr 5 '12 at 11:08
You miss a crucial condition on $C$ and how it must relate to $f$ in your statement of Cauchy's theorem... – t.b. Apr 5 '12 at 11:32
I'm returning to this problem. I can't determine from the text if $L$ in the theorem is length of the curve before or after the function is applied. IOW, if $f(z)=z^2$ and $C$ is $|z|=R$, then is $L=2 \pi R$ - the circumference of $C$ - or is $L=2 \pi R^2$ - the circumference of the curve after passing through $f$? – Jeff Apr 17 '12 at 21:07
Per @t.b., $L=2 \pi R$. – Jeff Apr 17 '12 at 22:02

1 Answer 1

up vote 1 down vote accepted


The idea is that you should use Cauchy's theorem to prove that the value of the integral does not depend on the value of $R \in (0,\infty).$ If you then prove that $|\int f| \leq ML \to 0$ as $R \to 0$ or $R \to \infty$, then you will have shown that the integral must vanish.

Also, you should take a second look at your answers to part two. (Your answer for $n = -1$ is wrong, and you missed the case $n=0$.)

share|cite|improve this answer
I fixed my answers to part two. I think I had more mistakes than just $n=-1$. But I'm still not quite seeing what happens here. Let's consider just when $n \geq 0$. In that case, the expression $2 \pi R^{n+1}$ converges to different values as $R \rightarrow \infty$ and $R \rightarrow 0$. How are we to show that the integral's value does not depend on $R$? Only for $n=-1$ does the limit not depend on $R$. – Jeff Apr 18 '12 at 21:25
You show independence of $R$ by applying Cauchy's theorem to the curve $C$ formed by two disjoint circles around the origin. The interior of the curve $C$ is an annulus, , where $z^n$ is certainly holomorphic. This shows that the two integrals over the two boundary curves are equal when counted with appropriate signs. – Dan Petersen Apr 19 '12 at 9:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.