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Let $f$ be holomorphic on a disk $|z-z_0| \leq r$ and let $|f(z)| \leq M$ for $z$ inside the disk. I want to show that $$\left \lvert \frac{1}{2 \pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}} \, dz \right \rvert \leq \frac{M}{r^n} $$

My attempt so far is below:

\begin{align} \left \lvert \frac{1}{2 \pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}} \, dz \right \rvert &= \frac{1}{2 \pi} \left \lvert \int_C \frac{f(z)}{(z-z_0)^{n+1}} \, dz \right \rvert \\ & \leq \frac{1}{2\pi} \cdot \frac{M}{|z-z_0|^{n+1}} \cdot 2\pi r\\ & = \frac{rM}{|z-z_0|^{n+1}} \end{align}

However, I'm not sure how to find a lower bound for the denominator in order to get the result. I assume the lower bound must be $r^{n+1}$, but why?

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  • $\begingroup$ Can you tell us what is $C$? I would like to be sure that $z_0$ is inside $C$. $\endgroup$ Dec 7, 2016 at 23:11
  • $\begingroup$ If $C$ winds around $z_0$ more than once, the estimate need not hold. So indeed, what is $C$? $\endgroup$ Dec 7, 2016 at 23:39

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I assume $C$ (traversing once) denotes the circle of radius $r$ about $z_0$. In this case, when integrating, $\frac{1}{2\pi}\int_C\frac{|f(z)|}{|z-z_0|^{n+1}}dz \leq \frac{1}{2\pi}\frac{M}{r^{n+1}}2\pi r = \frac{M}{r^n}$.

Edit:misread the question. I've updated the solution.

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