which of the following is NOT a possible value of $(e^{f})''(0)$??

Question

Let $f$ be an analytic function on $\bar{} = \{z \in \mathbb{C}: |z| \le 1\}$. Assume that $|()| ≤ 1$ for each $z\in \bar{D}$. Then, which of the following is NOT a possible value of $(e^{f})''(0)$??

$(A) 2$

$(B) 6$

$(C) \frac{7}{9}e^{\frac{1}{9}}$

$(D) √2 + √2$

So I take a look at Cauchy Integral Formula and then $$(e^{f})''(0)=\frac{1}{\pi}\int_{|z|=1}\frac{e^{f(z)}}{z^3}$$

Taking modulus on both sides we have $$|(e^{f})''(0)| \le \frac{e}{\pi}2\pi=2e$$

Hence (b) cannot be the choice.

Is this alright!!

Thanks for the help!

Answer 1

(B) can not be the choice, I agree.

On the contrary, all $\alpha $ with $|\alpha |\le 2$ are possible values of $(e^{f})^{\prime\prime}(0)$.
Let $f(z)=\frac{\alpha }{2}z^2$, then $f(z)$ satisfies the condition $|f(z)|\le 1$.
Furthermore $(e^{f})^{\prime\prime}(0)=\alpha $, since $(e^{f})^{\prime\prime}(z)=e^{f(z)}(\alpha ^2z^2+\alpha )$ and $f(0)=0$.
Therefore (A), (C) and (D) are possible values of $(e^{f})^{\prime\prime}(0)$.