• Title: Liouville’s Theorem

  • Series: Complex Analysis

  • YouTube-Title: Complex Analysis 29 | Liouville’s Theorem

  • Bright video: https://youtu.be/RfJ2bOfsdgY

  • Dark video: https://youtu.be/aG7qQcoElYU

  • Quiz: Test your knowledge

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  • Subtitle on GitHub: ca29_sub_eng.srt missing

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  • Quiz Content

    Q1: Let $f: D \rightarrow \mathbb{C}$ be a holomorphic function where $\overline{B_r(z_0)} \subseteq D$. What is the correct Cauchy’s inequality?

    A1: $$| f^\prime(z_0) | = \frac{1}{r} \sup_{ z \in \partial B_r(z_0)} | f(z) |$$

    A2: $$| f^\prime(z_0) | = \frac{1}{r^2} \sup_{ z \in \partial B_r(z_0)} | f(z) |$$

    A3: $$| f^\prime(z_0) | = \frac{1}{r^3} \sup_{ z \in \partial B_r(z_0)} | f(z) |$$

    A4: $$| f^\prime(z_0) | = \frac{4}{r^2} \sup_{ z \in \partial B_r(z_0)} | f(z) |$$

    Q2: Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function given by $f(z) = \cos(z)$. Is there a $z_1$ with $|\cos(z_1)| \geq 10$?

    A1: Yes, there is.

    A2: Yes, but $z \in \mathbb{R}$.

    A3: No, $\cos$ is bounded.

    Q3: Let $f: D \rightarrow \mathbb{C}$ be a holomorphic function that is bounded and not constant. Is it possible that $D = \mathbb{C}$?

    A1: No, that is not possible.

    A2: Yes, there are a lot of examples.

  • Last update: 2024-10

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