# [SOLVED]meromorphic

#### dwsmith

##### Well-known member
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?).

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#### Opalg

##### MHB Oldtimer
Staff member
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 $\color{blue}R/n<1/2.$
...

• dwsmith

#### dwsmith

##### Well-known member
Since the summation is only cycling through positive values, how would n every be negative?

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

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#### Opalg

##### MHB Oldtimer
Staff member
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.

• dwsmith