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Chapter 8: Problem 12

Let \(f(z)\) be analytic in \(|z|

Short Answer

Expert verified
Turns out, if function \(f(z)\) is analytic in \(|z|<R\), the function \(F(z) = f(z) + f(z^2) + f(z^3) + ...\), can also be expressed as a power series, thus is also analytic in \(|z|<R\). By applying the Maximum Modulus Principle and the triangle inequality to \(F(z)\), it can be shown that \(|F(z)| \leq \frac{r}{1-r} \quad(|z|=r<R)\).

Step by step solution

01

Assumption of Analyticity

We start from the assumption that \(f(z)\) is analytic in \(|z|<R\). By definition, an analytic function can be represented as a power series. Thus, the function \(f(z)\) can be written as: \(f(z) = a_1z+a_2z^2+...+a_nz^n+...\) within the radius of convergence \(|z|<R\).
02

Expressing Function F(z)

Next, let's express \(F(z)\) as a power series. \(F(z)=f(z)+f\left(z^{2}\right)+f\left(z^{3}\right)+\cdots\) can be written as an infinite sum of \(f(z^n)\). We can substitute the power series we've found for \(f(z)\). Then, by multiplying out the terms and reordering them, we can express \(F(z)\) as its own power series.
03

Confirming Analyticity

In step 2, if we can successfully express \(F(z)\) as a power series, it means \(F(z)\) is analytic within the same radius of convergence. This is in line with the definition of an analytic function, which can be represented as a power series inside its radius of convergence.
04

Applying Maximum Modulus Principle

To show \(|F(z)| \leq \frac{r}{1-r} \quad(|z|=r<R)\), we can apply Maximum Modulus Principle which says the maximum magnitude of an analytic function within a region is taken on its boundary. From the triangle inequality, we can find a bound for \(|F(z)|\). By simplifying, we can get the desired inequality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series
A power series is an infinite series of the form \(\sum_{n=0}^\infty a_n(z-c)^n\), where \(a_n\) are coefficients, \(z\) is the variable, and \(c\) is the center of the series. This representation is central to understanding analytic functions, such as \(f(z)\) described in the original exercise.

Each term \(a_n(z-c)^n\) is a product of a coefficient and \(z-c\) raised to the power of \(n\), which allows this format to express a wide variety of functions within a certain range from \(c\). The power series is particularly useful because it can be differentiated and integrated term-by-term within its radius of convergence.

One reason mathematicians love power series is their simplicity in manipulation. They can be added, subtracted, multiplied, and often reorganized, as shown in the step-by-step solution where the function \(F(z)\) was created from the function \(f(z)\) and its powers.
Radius of Convergence
The radius of convergence is a measure that tells us where a power series converges to the function it represents. Specifically, for a power series centered at \(c\), it converges on the disk \(|z-c|
The determination of this radius is a critical step in the analysis of power series and analytic functions. It can be determined using various methods, including the Ratio Test or Root Test. Once you know the radius of convergence, it becomes possible to understand the behavior of the function represented by the power series within that disk.

In our given exercise example, knowing that \(f(z)\) is analytic inside the disk \(|z|
Maximum Modulus Principle
The Maximum Modulus Principle is a profound concept in complex analysis that deals with the behaviour of analytic functions. This principle states that if a function is analytic and non-constant within a given domain, the function's magnitude will reach its maximum on the boundary of that domain.

This principle is used to determine bounds for the size of a function within a given region. In the context of the exercise, the Maximum Modulus Principle helps us establish the inequality \(|F(z)| \leq \frac{r}{1-r}\) when \(|z|=r
Understanding this principle is essential as it relates the values that an analytic function can take within a domain to the values it takes at the boundary and it is a powerful tool in complex analysis, impacting other conclusions drawn about analytic functions such as \(F(z)\) in our exercise.

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Most popular questions from this chapter

$$ \begin{aligned} &\text { Prove that there is no analytic function } f \text { in the unit disk } \Delta=\\\ &\\{z:|z|<1\\} \text { such that } f(1 / n)=(-1)^{n} / n^{2} \text { for } n=2,3,4, \ldots \end{aligned} $$

Let \(f\) and \(g\) be analytic in the unit disk \(\Delta\). (a) If \(f(1 / n)=g(1 / n)\) for \(n=2,3, \ldots\), show that \(f=g\). (b) Show that \(f(1 / n)=1 / \sqrt{n}\) for each \(n=2,3, \ldots\) is not possible.

If \(f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}\) is analytic in \(|z|

Find the first five coefficients in the Maclaurin expansion for (a) \(e^{z} \sin z\) (b) \(\frac{1}{\cos z}\) (c) \(e^{z+z^{2}}\) (d) \(e^{z /(1-z)}\).

Suppose that \(f(z)\) is an entire function such that \(\left|f^{\prime}(z)\right| \leq|z|\) for all \(z \in \mathbb{C} .\) Show that \(f\) must be of the form \(f(z)=a z^{2}+b\) where \(a, b\) are complex constants such that \(|a| \leq 1 / 2\). What will be the form of \(f\) if \(f(z)\) is entire such that \(\left|f^{(k)}(z)\right| \leq|z|\) for some fixed \(k>2\) and for all \(z \in \mathbb{C} ?\)
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