The question is: if $f(z)$ is analytic and $|f(z)|\leq M$ for $|z|\leq r$, find an upper bound for$|f^{(n)}(z)|$ in $|z|\leq\frac{r}{2}$. My attempt: Since $f(z)$ is analytic where $|z|\leq r$, we know that $$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw,$$ and $f(z)$ is bounded by $M$. We know that $\bigg|\int_Cf(z)dz\bigg|\leq\max|f(z)|\cdot\text{(length of C)}$, so $$\bigg|f^{(n)}(z)\bigg|=\bigg|\frac{n!}{2\pi i}\int_{|z|=r}\frac{f(w)}{(w-z)^{n+1}}dw\bigg|\leq n!\cdot M_n\cdot2\pi r,$$ where $M_n:=\max|\frac{f(w)}{(w-z)^{n+1}}|$, for […]

Let $f$ be an analytic function on $\bar{𝐷} = \{z \in \mathbb{C}: |z| \le 1\}$. Assume that $|𝑓(𝑧)| ≤ 1$ for each $z\in \bar{D}$. Then, which of the following is NOT a possible value of $(e^{f})”(0)$?? $(A) 2$ $(B) 6$ $(C) \frac{7}{9}e^{\frac{1}{9}}$ $(D) √2 + 𝑖√2$ So I take a look at Cauchy Integral Formula […]

Given an Annulus with $A(0,r,R)$ show by considering Cauchy’s Theorem for primitives that there is no holomorphic function with $f'(z)=\dfrac{1}{z}$. I am struggling to picture this since but it seems like there are issues because $f(z)=\log z$ isn’t well defined in the same range as $\dfrac{1}{z}$. Am I looking to show that $\int_Ag(z)dz = 0$ […]

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