Let $\mu$ be the Borel probability measured on the open unit disk $\mathbb{D}= \{z \in \mathbb{C}: |z| < 1\}.$
Take \begin{equation*} f_n(z) = \sum_{k= 2^n}^{2^{n+1}-1} z^k = z^{2^n}\frac{1 - z^{2^n}}{1-z} \end{equation*} and define \begin{equation*} R_n := \{ z \in \mathbb{D}: 1 - 1/2^n \leq |z| < 1 - 1/2^{n+1} \text{and} \ \arg(z) \in (2 \pi j/2^n, 2 \pi (j+1)/2^n \}. \end{equation*} I want to prove that there exists $C > 0$ such that \begin{equation*} \left| \int_{R_n} f_n(z) d\mu(z) \right| \geq C ( 2^n \mu(R_n) )^{1/2}. \end{equation*} I proved that exists $\delta > 0$ such that $|f_n(z)| \geq \delta 2^{2n} \mu(R_{n})$ and I try to integrate, but I'm not able to get the modulus outside the integral. How do I continue?