AI Summary & Key Takeaways
AI InsightMoreover, we prove that $l(r,f)=\mathcal{O}\left( \log\frac{1}{1-r} \right)$ and that $L(r,f)=\mathcal{O}\left\{ A(r,f)\log \frac{1}{1-r} \right\}^{1/2}$ as $r\to 1$ under weaker assumptions on $f(z)$ than some previous results of this type.
📌 3 Key Takeaways
Abstract
Let $\mathcal{H}$ denote the class of analytic functions in the unit disk $|z|<1$. Let $C(r,f)$ be the closed curve which is the image of the circle $|z|=r<1$ under the mapping $w=f(z)\in\mathcal{H}$, $L(r,f)$ the length of $C(r,f)$ and let $A(r,f)$ be the area enclosed by $C(r,f)$. Let $l(re^{i\theta},f)$ be the length of the image curve of the line segment joining $re^{i\theta}$ and $re^{i(\theta+\pi)}$ under the mapping $w=f(z)$ and let $l(r,f)=\max_{0\leq\theta 2 \pi}l(re^{i\theta},f)$. We find upper bound for $l(r,f)$ for $f(z)$ in some known classes of functions. Moreover, we prove that $l(r,f)=\mathcal{O}\left( \log\frac{1}{1-r} \right)$ and that $L(r,f)=\mathcal{O}\left\{ A(r,f)\log \frac{1}{1-r} \right\}^{1/2}$ as $r\to 1$ under weaker assumptions on $f(z)$ than some previous results of this type.