Post History
#4: Post edited
by
Snoopy
·
2024-08-16T02:17:10Z (3 months ago)
- > **Problem**. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_0 \in \Omega$ such that
- > $$
- \varphi\left(z_0\right)=z_0 \quad \text { and } \quad \varphi^{\prime}\left(z_0\right)=1
- $$
- > then $\varphi$ is linear.
- This exercise (Exercise 9 in Chapter 2) from [*Complex Analysis* by Stein-Shakarchi](https://press.princeton.edu/books/hardcover/9780691113852/complex-analysis) is intended to be solved using Cauchy's Theorem or its corollaries presented in the chapter.
- However, the authors omitted the assumption that $\Omega$ is connected. This assumption is necessary, as a counterexample can easily be constructed by defining $\varphi(z) = z^2$ on the connected components that do not contain $z_0$.
- The textbook provides a critical hint for solving the problem:
- > Hint: Why can one assume that $z_0=0$ ? Write $\varphi(z)=z+a_n z^n+O\left(z^{n+1}\right)$ near 0 , and prove that if $\varphi_k=\varphi \circ \cdots \circ \varphi$ (where $\varphi$ appears $k$ times), then $\varphi_k(z)=$ $z+k a_n z^n+O\left(z^{n+1}\right)$. Apply the Cauchy inequalities and let $k \rightarrow \infty$ to conclude the proof. Here we use the standard $O$ notation, where $f(z)=O(g(z))$ as $z \rightarrow 0$ means that $|f(z)| \leq C|g(z)|$ for some constant $C$ as $|z| \rightarrow 0$.
- **Comments on the hint:**
- - The reduction to assuming $z_0 = 0$ is a fairly standard step for experienced readers but may appear mysterious to beginners. Understanding the details of "WLOG" (Without Loss of Generality) part in various arguments is rewarding.
- The hint mentions the standard O notation in its asymptotic form. In number theory, a non-asymptotic form, where the inequality $|f(z)| \le Cg(z)$ holds for *all* $z$ in the domain, is frequently used.- ---
- I will write my solution to the problem below. Alternative approaches or broader insights beyond the original question are also welcome.
- > **Problem**. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_0 \in \Omega$ such that
- > $$
- \varphi\left(z_0\right)=z_0 \quad \text { and } \quad \varphi^{\prime}\left(z_0\right)=1
- $$
- > then $\varphi$ is linear.
- This exercise (Exercise 9 in Chapter 2) from [*Complex Analysis* by Stein-Shakarchi](https://press.princeton.edu/books/hardcover/9780691113852/complex-analysis) is intended to be solved using Cauchy's Theorem or its corollaries presented in the chapter.
- However, the authors omitted the assumption that $\Omega$ is connected. This assumption is necessary, as a counterexample can easily be constructed by defining $\varphi(z) = z^2$ on the connected components that do not contain $z_0$.
- The textbook provides a critical hint for solving the problem:
- > Hint: Why can one assume that $z_0=0$ ? Write $\varphi(z)=z+a_n z^n+O\left(z^{n+1}\right)$ near 0 , and prove that if $\varphi_k=\varphi \circ \cdots \circ \varphi$ (where $\varphi$ appears $k$ times), then $\varphi_k(z)=$ $z+k a_n z^n+O\left(z^{n+1}\right)$. Apply the Cauchy inequalities and let $k \rightarrow \infty$ to conclude the proof. Here we use the standard $O$ notation, where $f(z)=O(g(z))$ as $z \rightarrow 0$ means that $|f(z)| \leq C|g(z)|$ for some constant $C$ as $|z| \rightarrow 0$.
- **Comments on the hint:**
- - The reduction to assuming $z_0 = 0$ is a fairly standard step for experienced readers but may appear mysterious to beginners. Understanding the details of "WLOG" (Without Loss of Generality) part in various arguments is rewarding.
- - The hint mentions the standard big O notation in its asymptotic form. In number theory, a non-asymptotic form, where the inequality $|f(z)| \le C|g(z)|$ holds for *all* $z$ in the domain, is frequently used.
- ---
- I will write my solution to the problem below. Alternative approaches or broader insights beyond the original question are also welcome.
#3: Post edited
by
Snoopy
·
2024-08-15T23:34:08Z (3 months ago)
holomorphic function on a connected bounded open subset of $\mathbf{C}$
- Holomorphic function on a connected bounded open subset of the complex plane
#2: Post edited
by
Snoopy
·
2024-08-15T21:33:10Z (3 months ago)
- > **Problem**. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_0 \in \Omega$ such that
- > $$
- \varphi\left(z_0\right)=z_0 \quad \text { and } \quad \varphi^{\prime}\left(z_0\right)=1
- $$
- > then $\varphi$ is linear.
This exercise (Exercise 9 in Chapter 2) from *Complex Analysis* by Stein-Shakarchi is intended to be solved using Cauchy's Theorem or its corollaries presented in the chapter.- However, the authors omitted the assumption that $\Omega$ is connected. This assumption is necessary, as a counterexample can easily be constructed by defining $\varphi(z) = z^2$ on the connected components that do not contain $z_0$.
- The textbook provides a critical hint for solving the problem:
- > Hint: Why can one assume that $z_0=0$ ? Write $\varphi(z)=z+a_n z^n+O\left(z^{n+1}\right)$ near 0 , and prove that if $\varphi_k=\varphi \circ \cdots \circ \varphi$ (where $\varphi$ appears $k$ times), then $\varphi_k(z)=$ $z+k a_n z^n+O\left(z^{n+1}\right)$. Apply the Cauchy inequalities and let $k \rightarrow \infty$ to conclude the proof. Here we use the standard $O$ notation, where $f(z)=O(g(z))$ as $z \rightarrow 0$ means that $|f(z)| \leq C|g(z)|$ for some constant $C$ as $|z| \rightarrow 0$.
- **Comments on the hint:**
- - The reduction to assuming $z_0 = 0$ is a fairly standard step for experienced readers but may appear mysterious to beginners. Understanding the details of "WLOG" (Without Loss of Generality) part in various arguments is rewarding.
- - The hint mentions the standard O notation in its asymptotic form. In number theory, a non-asymptotic form, where the inequality $|f(z)| \le Cg(z)$ holds for *all* $z$ in the domain, is frequently used.
- ---
- I will write my solution to the problem below. Alternative approaches or broader insights beyond the original question are also welcome.
- > **Problem**. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_0 \in \Omega$ such that
- > $$
- \varphi\left(z_0\right)=z_0 \quad \text { and } \quad \varphi^{\prime}\left(z_0\right)=1
- $$
- > then $\varphi$ is linear.
- This exercise (Exercise 9 in Chapter 2) from [*Complex Analysis* by Stein-Shakarchi](https://press.princeton.edu/books/hardcover/9780691113852/complex-analysis) is intended to be solved using Cauchy's Theorem or its corollaries presented in the chapter.
- However, the authors omitted the assumption that $\Omega$ is connected. This assumption is necessary, as a counterexample can easily be constructed by defining $\varphi(z) = z^2$ on the connected components that do not contain $z_0$.
- The textbook provides a critical hint for solving the problem:
- > Hint: Why can one assume that $z_0=0$ ? Write $\varphi(z)=z+a_n z^n+O\left(z^{n+1}\right)$ near 0 , and prove that if $\varphi_k=\varphi \circ \cdots \circ \varphi$ (where $\varphi$ appears $k$ times), then $\varphi_k(z)=$ $z+k a_n z^n+O\left(z^{n+1}\right)$. Apply the Cauchy inequalities and let $k \rightarrow \infty$ to conclude the proof. Here we use the standard $O$ notation, where $f(z)=O(g(z))$ as $z \rightarrow 0$ means that $|f(z)| \leq C|g(z)|$ for some constant $C$ as $|z| \rightarrow 0$.
- **Comments on the hint:**
- - The reduction to assuming $z_0 = 0$ is a fairly standard step for experienced readers but may appear mysterious to beginners. Understanding the details of "WLOG" (Without Loss of Generality) part in various arguments is rewarding.
- - The hint mentions the standard O notation in its asymptotic form. In number theory, a non-asymptotic form, where the inequality $|f(z)| \le Cg(z)$ holds for *all* $z$ in the domain, is frequently used.
- ---
- I will write my solution to the problem below. Alternative approaches or broader insights beyond the original question are also welcome.
#1: Initial revision
by
Snoopy
·
2024-08-15T21:25:56Z (3 months ago)
holomorphic function on a connected bounded open subset of $\mathbf{C}$
> **Problem**. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_0 \in \Omega$ such that
> $$
\varphi\left(z_0\right)=z_0 \quad \text { and } \quad \varphi^{\prime}\left(z_0\right)=1
$$
> then $\varphi$ is linear.
This exercise (Exercise 9 in Chapter 2) from *Complex Analysis* by Stein-Shakarchi is intended to be solved using Cauchy's Theorem or its corollaries presented in the chapter.
However, the authors omitted the assumption that $\Omega$ is connected. This assumption is necessary, as a counterexample can easily be constructed by defining $\varphi(z) = z^2$ on the connected components that do not contain $z_0$.
The textbook provides a critical hint for solving the problem:
> Hint: Why can one assume that $z_0=0$ ? Write $\varphi(z)=z+a_n z^n+O\left(z^{n+1}\right)$ near 0 , and prove that if $\varphi_k=\varphi \circ \cdots \circ \varphi$ (where $\varphi$ appears $k$ times), then $\varphi_k(z)=$ $z+k a_n z^n+O\left(z^{n+1}\right)$. Apply the Cauchy inequalities and let $k \rightarrow \infty$ to conclude the proof. Here we use the standard $O$ notation, where $f(z)=O(g(z))$ as $z \rightarrow 0$ means that $|f(z)| \leq C|g(z)|$ for some constant $C$ as $|z| \rightarrow 0$.
**Comments on the hint:**
- The reduction to assuming $z_0 = 0$ is a fairly standard step for experienced readers but may appear mysterious to beginners. Understanding the details of "WLOG" (Without Loss of Generality) part in various arguments is rewarding.
- The hint mentions the standard O notation in its asymptotic form. In number theory, a non-asymptotic form, where the inequality $|f(z)| \le Cg(z)$ holds for *all* $z$ in the domain, is frequently used.
---
I will write my solution to the problem below. Alternative approaches or broader insights beyond the original question are also welcome.