Exploring Problem 167 in Serge Lang's Complex Analysis

[Dustinsfl]

I am trying to understand a problem in my book (for reference pr 167 Serge Lang Complex Analysis).

$$
f(z) = \frac{1}{z} + \sum_{n = 1}^{\infty}\frac{z}{z^2-n^2}
$$

Let R>0 (is this R representing the radius of convergence?) and let N>2R (where did this come from and why?).

Write $f(z) = g(z)+h(z)$ where
$$
g(z) = \frac{1}{z}+\sum_{n = 1}^{N}\frac{z}{z^2-n^2} \quad\text{and}\quad
h(z) = \sum_{N+1}^{\infty}\frac{z}{z^2-n^2}
$$
g is a rational function and is meromorphic on C (why are rational functions automatically meromorphic?).
We see that g has simple poles at the integers n such that $|n|\leq N$ (why are we looking at the absolute value of n?)

For $|z|<R$ we have the estimate
$$
\left|\frac{z}{z^2-n^2}\right|\leq\frac{R}{n^2-R^2}=\frac{1}{n^2}\frac{R}{1-\left(R/n)\right)^2}
$$
(\frac{R}{n^2-R^2} why is that?)

The denominator satisfies $1-\left(R/n\right)^2\geq 3/4$ for $n>N>2R$ (Why is this?).

 

[Opalg]

I am trying to understand a problem in my book (for reference pr 167 Serge Lang Complex Analysis).

$$
f(z) = \frac{1}{z} + \sum_{n = 1}^{\infty}\frac{z}{z^2-n^2}
$$

Let R>0 (is this R representing the radius of convergence?) and let N>2R (where did this come from and why?).
R is not a radius of convergence, it is a positive constant whose significance will only emerge later in the proof. Ditto for N.

Write $f(z) = g(z)+h(z)$ where
$$
g(z) = \frac{1}{z}+\sum_{n = 1}^{N}\frac{z}{z^2-n^2} \quad\text{and}\quad
h(z) = \sum_{N+1}^{\infty}\frac{z}{z^2-n^2}
$$
g is a rational function and is meromorphic on C (why are rational functions automatically meromorphic?).
A rational function is analytic everywhere except where its denominator is zero. The denominator is a polynomial, so it has only finitely many zeros, each of which is a pole of g. Thus g is analytic except at finitely many poles, and is therefore meromorphic.

We see that g has simple poles at the integers n such that $|n|\leq N$ (why are we looking at the absolute value of n?)
The poles of g occur at points where $\color{blue}z^2-n^2=0$ for some n with $\color{blue}n\leqslant N.$ Thus $\color{blue}z=\pm n.$ So the poles occur at negative as well as positive points on the real axis.

For $|z|<R$ we have the estimate
$$
\left|\frac{z}{z^2-n^2}\right|\leq\frac{R}{n^2-R^2}=\frac{1}{n^2}\frac{R}{1-\left(R/n)\right)^2}
$$
(\frac{R}{n^2-R^2} why is that?)
It doesn't say so, but at this stage we must be assuming that $\color{blue}|n|>N.$ Thus $\color{blue}n^2>N^2>R^2>|z|^2$, and so $\color{blue}|z^2-n^2| > n^2-R^2$ (triangle inequality).

The denominator satisfies $1-\left(R/n\right)^2\geq 3/4$ for $n>N>2R$ (Why is this?).
That follows trivially from the fact that $\color{blue}R/n<1/2.$

...

 

[Dustinsfl]

Since the summation is only cycling through positive values, how would n every be negative?

So for the triangle inequality, $z^2 < R^2\iff z^2-n^2+n^2<R^2\iff z^2-n^2<R^2-n^2\iff |n^2-z^2|<|n^2-R^2|$.

 

[Opalg]

Since the summation is only cycling through positive values, how would n every be negative?

The reason this is confusing is that $n$ is being used in two different senses. In the summation, $n$ is a positive integer. But when indicating the points where $g$ has a pole, it also takes negative values.

 

[blue_raver22]

In response to your question, Problem 167 in Serge Lang's Complex Analysis deals with the function $f(z)$, which is defined as $\frac{1}{z} + \sum_{n=1}^{\infty} \frac{z}{z^2-n^2}$. The goal of this problem is to understand the behavior of this function and its properties.

First, let's address the notation used in the problem. The letter $R$ in this context is representing the radius of convergence, which is the radius of the largest circle centered at the origin within which the series will converge. The letter $N$ is simply a variable used to represent any number greater than $2R$, which will be explained in the following steps.

Now, let's break down the problem and understand the reasoning behind each step. The first step is to write $f(z)$ as the sum of two functions, $g(z)$ and $h(z)$, where $g(z)$ is a rational function and $h(z)$ is a series. This will help us understand the properties of $f(z)$ better.

Next, we see that $g(z)$ has simple poles at the integers $n$ such that $|n|\leq N$. This means that $g(z)$ will have poles at $n=\pm 1, \pm 2, ..., \pm N$. This is because the denominator of $g(z)$ will be equal to zero at these points, making the function undefined.

Now, we come to the main part of the problem, which is estimating the function $f(z)$ for $|z|<R$. The estimate given in the problem is $|\frac{z}{z^2-n^2}|\leq \frac{R}{n^2-R^2}$. This estimate is based on the fact that for $|z|<R$, the terms in the series will become smaller and smaller as $n$ increases. This is because as $n$ increases, the denominator of the terms, which is $z^2-n^2$, will also increase, making the overall fraction smaller.

The denominator in the estimate, $1-(R/n)^2$, satisfies the inequality $1-(R/n)^2 \geq 3/4$ for $n>N>2R$. This is because for $n>N>2R$, the value of $R