3
$\begingroup$

I'm currently revising some complex analysis, and need to show that the series $$\sum_{n=-\infty}^{\infty}\frac{1}{n^2 - z^2}$$ defines a holomorphic function on $\mathbb{C}\setminus\mathbb{Z}$. The hint that the question gives me is to consider any disc contained in $\mathbb{C}\setminus\mathbb{Z}$ and show that the series converges uniformly there.

I can see how to complete the question once I've done this, but it's late, I'm tired, and I can't figure out how to show the series converges uniformly. I've tried the Weierstrass M-Test, but this failed for me on discs where $z$ was close to $n$, for obvious reasons.

Any hints would be greatly appreciated.

$\endgroup$
1
  • $\begingroup$ Something involving Laurent Series/Cauchy Integral formula? Not sure otherwise $\endgroup$ – Chill2Macht May 21 '16 at 22:51
3
$\begingroup$

Let $K \subset (B(0,R) \cap \mathbb{C} \setminus \mathbb{Z})$ be a compact set and set $d(K,\mathbb{Z}) = \delta > 0$. If $|n| > R$, then for all $x \in K$ we have

$$ |n^2 - x^2| \geq ||n|^2 - |x|^2| = |n|^2 - |x|^2 \geq |n|^2 - R^2. $$

If $|n| \leq R$ then for all $x \in K$ we have

$$ |n^2 - x^2| = |x - n||x - (-n)| \geq \delta^2. $$

Hence,

$$ \sup_{x \in K} \left| \frac{1}{n^2 - x^2} \right| \leq \begin{cases} \frac{1}{\delta^2} & |n| \leq R, \\ \frac{1}{|n|^2 - R^2} & |n| > R \end{cases} $$

and so the series converges uniformly on $K$ by the Weierstrass $M$-test. Finish by taking $K = \overline{B(x_0,r)}$ for all $x_0 \in \mathbb{C} \setminus \mathbb{Z}$ and small enough $r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.