Question

(a) If $f(z)$ is analytic inside and on the simple closed curve $C$ enclosing $z=a$, prove that $$ \{f(a)\}^n=\frac{1}{2 \pi i} \oint_C \frac{\{f(z)\}^n}{z-a} d z \quad n=0,1,2, \ldots $$ (b) Use (a) to prove that $|f(a)|^n \leq M^n / 2_\pi D$ where $D$ is the minimum distance from $a$ to the curve $C$ and $M$ is the maximum value of $|f(z)|$ on $C$. (c) By taking the $n$th root of both sides of the inequality in (b) and letting $n \rightarrow \infty$, prove the maximum modulus theorem.

   (a) If $f(z)$ is analytic inside and on the simple closed curve $C$ enclosing $z=a$, prove that
$$
\{f(a)\}^n=\frac{1}{2 \pi i} \oint_C \frac{\{f(z)\}^n}{z-a} d z \quad n=0,1,2, \ldots
$$
(b) Use (a) to prove that $|f(a)|^n \leq M^n / 2_\pi D$ where $D$ is the minimum distance from $a$ to the curve $C$ and $M$ is the maximum value of $|f(z)|$ on $C$.
(c) By taking the $n$th root of both sides of the inequality in (b) and letting $n \rightarrow \infty$, prove the maximum modulus theorem.

Show more…

Schaum's Outlines: Complex Variables

Schaum's Outlines: Complex Variables

Murray Spiegel 1st Edition

Chapter 5, Problem 57

Instant Answer

Step 1

\] Show more…

Show all steps

lock

Thanks for your feedback!

(a) If $f(z)$ is analytic inside and on the simple closed curve $C$ enclosing $z=a$, prove that $$ \{f(a)\}^n=\frac{1}{2 \pi i} \oint_C \frac{\{f(z)\}^n}{z-a} d z \quad n=0,1,2, \ldots $$ (b) Use (a) to prove that $|f(a)|^n \leq M^n / 2_\pi D$ where $D$ is the minimum distance from $a$ to the curve $C$ and $M$ is the maximum value of $|f(z)|$ on $C$. (c) By taking the $n$th root of both sides of the inequality in (b) and letting $n \rightarrow \infty$, prove the maximum modulus theorem.

Play audio

Feedback

Powered by NumerAI

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later