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 $C$.
- Use (1) to give a bound for the integral of $f(z) = z^n$ ($n$ an integer, possibly negative) on the circle $\left|z\right| = R$.
- 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?)
- 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 $\int_C f(z)\ dz=0.$
- is easy: the integral $ML = 2 \pi R^{n+1}$.
- 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?