Equality holding in Cauchy's Inequality

Hello, this was a problem on an exam:

From Cauchy's Integral Formula, we get Cauchy's Inequality $\displaystyle |f^{(n)}(z_0)| \leq \frac{n!M}{R^n}$.

Prove that equality holds if and only if $\displaystyle f(z)= \frac{aMz^n}{R^n}$ for some $\displaystyle a \in \mathbb{C}$ with $\displaystyle |a| = 1$.

Proving $\displaystyle (\Leftarrow)$ is easy, just take the modulus of the nth derivative of f.

How do we prove the other direction?

Thanks!

Re: Equality holding in Cauchy's Inequality

Re: Equality holding in Cauchy's Inequality

Ah, From Cauchy's Integral Formula and Inequality.

f is analytic on a ball of radius R

|f(z)| __<__ M for all z in the ball

Re: Equality holding in Cauchy's Inequality

So the statement is:

Let $\displaystyle \Omega = \{z \in \mathbb{C} \ | \ \lVert z - z_0 \rVert < R \}$. Let $\displaystyle f \in H(\bar{\Omega})$ satisfy $\displaystyle \lVert f(z) \rVert \le M \ \forall z \in \Omega$.

Then $\displaystyle \lVert f^{(n)}(z_0) \rVert = \frac{n!M}{R^n} \Leftrightarrow \exists a \in \mathbb{C}, \lVert a \rVert = 1$, such that $\displaystyle f(z) = \frac{aMz^n}{R^n} \ \forall z \in \Omega$.

Correct?

Are you sure it isn't supposed to be $\displaystyle f(z) = \frac{aM(z-z_0)^n}{R^n}$ ?

Re: Equality holding in Cauchy's Inequality

What I wrote is what was on the exam, but I think that was justified by assuming that z_{0} = 0.

And yes, what you have is correct, as far as I understood the problem