Bourbaki

Exercise 1, § 3

General Topology, VIII

For every finite sequence s=(zk)k∈Is = (z_k)_{k \in I} of complex numbers, put

∑k∈I∣zk∣=ρs⋅sup⁡J⊂I∣∑k∈Jzk∣ \sum_{k \in I} |z_k| = \rho_s \cdot \sup_{J \subset I} \left| \sum_{k \in J} z_k \right|
(cf. Chapter VII, § 3, no. 1, Proposition 2).

a) \* If zk=rk(cos⁡φk+isin⁡φk)z_k = r_k (\cos \varphi_k + i \sin \varphi_k), where rk=∣zk∣r_k = |z_k| and φk\varphi_k is the principal radian measure of the amplitude of zkz_k, put

f(φ)=∑k∈Irk(cos⁡(φ−φk))+ f(\varphi) = \sum_{k \in I} r_k (\cos (\varphi - \varphi_k))^+
for every φ∈[0,2π[\varphi \in [0, 2\pi[. Show that the least upper bound of f(φ)f(\varphi) in the interval [0,2π[[0, 2\pi[ is equal to sup⁡J⊂I∣∑k∈Jzk∣\sup_{J \subset I} \left| \sum_{k \in J} z_k \right|. By considering the integral
∫02πf(φ) dφ, \int_0^{2\pi} f(\varphi) \, d\varphi,
deduce that for all finite sequences ss we have ρs≤π\rho_s \leq \pi, and that there is no finite sequence (zk)=s(z_k) = s such that ρs=π\rho_s = \pi.

b) Show that, for every ε>0\varepsilon > 0, there exists a finite sequence (zk)=s(z_k) = s such that ρs>π−ε\rho_s > \pi - \varepsilon (take the zkz_k to be the roots of a binomial equation of sufficiently high degree). \*

Solution

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Transcribed from the printed volume. Not translated.