Uniform convergence

Complex Analysis
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Papapetros Vaggelis
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Uniform convergence

#1

Post by Papapetros Vaggelis »

Examine if there exists a sequence \(\displaystyle{\left(p_n(z)\right)_{n\in\mathbb{N}}}\) of

complex polynomials such that \(\displaystyle{p_n(z)\to \dfrac{1}{z}}\) uniformly to

\(\displaystyle{C_{r}=\left\{z\in\mathbb{C}: |z|=r\right\}}\) (\(\displaystyle{r>0}\)).
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Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Uniform convergence

#2

Post by Riemann »

Suppose that this is not the case. If such sequence existed then the convergence at the compact set $\left| z \right| = r$ would be uniform. However,

$$2 \pi i = \oint \limits_{\mathcal{C}_r} \frac{{\rm d}z}{z} = \oint \limits_{\mathcal{C}_r} \lim_{n \rightarrow +\infty} p_n (z) \, {\rm d}z = \lim_{n \rightarrow +\infty} \oint \limits_{\mathcal{C}_r} p_n (z) \, {\rm d}z = \lim_{n \rightarrow +\infty} 0 = 0 $$

since the polynomials have no poles inside the circle thus the contour integral is zero. The last relation is an obscurity of course. Thus, no complex polynomials exist. This completes our arguement.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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