# Cauchy's Theorem on $z^n$

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?

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Maybe you should use this variant of Cauchy's Theorem: en.wikipedia.org/wiki/Cauchy%27s_integral_formula –  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

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 $n=0$.)
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, en.wikipedia.org/wiki/Annulus_(mathematics) , 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