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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?).

$$

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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