Theorem 1 -- the strict one-half lower bound

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Statement

Let $n\geq1$ and let $z_1,\ldots,z_n\in\mathbb C$ satisfy $z_1=1$. For $j\geq1$, set

$$S_j=\sum_{t=1}^n z_t^j.$$

Then

$$\max_{1\leq j\leq n}|S_j|>\frac12. \tag{1}$$

Consequently the question in Problem 519 has an affirmative answer with the absolute constant $c=1/2$.

Newton--Girard interface

We use the following standard algebraic identity. If

$$P(x)=\prod_{t=1}^{m}(x-r_t) =x^m+c_1x^{m-1}+\cdots+c_m$$

and $p_j=\sum_{t=1}^{m}r_t^j$, then

$$p_k+c_1p_{k-1}+\cdots+c_{k-1}p_1+kc_k=0 \quad(1\leq k\leq m), \tag{2}$$

while

$$p_k+c_1p_{k-1}+\cdots+c_mp_{k-m}=0 \quad(k>m). \tag{3}$$

Equations (2)--(3), with their displayed sign convention, are the complete external Newton--Girard input used below.

If $n=1$, then $S_1=1$ and (1) is immediate. Assume henceforth that $n\geq2$, and write

$$\prod_{t=2}^{n}(x-z_t) =x^{n-1}+b_1x^{n-2}+\cdots+b_{n-1}.$$

Put

$$T_j=\sum_{t=2}^{n}z_t^j=S_j-1.$$

Applying (2) with $m=n-1$ gives, for $1\leq k\leq n-1$,

$$T_k+b_1T_{k-1}+\cdots+b_{k-1}T_1+kb_k=0.$$

After substituting $T_j=S_j-1$ and collecting constants, this becomes

$$S_k+b_1S_{k-1}+\cdots+b_{k-1}S_1 =1+b_1+\cdots+b_{k-1}-kb_k. \tag{4}$$

Applying (3) at $k=n=m+1$ gives

$$T_n+b_1T_{n-1}+\cdots+b_{n-1}T_1=0,$$

and therefore

$$S_n+b_1S_{n-1}+\cdots+b_{n-1}S_1 =1+b_1+\cdots+b_{n-1}. \tag{5}$$

These are equations (1) and (2), respectively, in the source.

Proof of the bound

Fix $0<\alpha<\pi/2$, and set

$$A_0=1, \qquad A_k=1+b_1+\cdots+b_k, \qquad W_k=1+|b_1|+\cdots+|b_k|.$$

For each $1\leq k\leq n-1$, the [[analysis/biro_1994_problem_turan_concerning_sums_powers_complex/lemma_1|geometric dichotomy]] says that either

$$|A_{k-1}-kb_k|\geq\sin\alpha\,|A_{k-1}|, \tag{6}$$

or

$$|A_k|\geq|A_{k-1}|+\cos\alpha\,|b_k|. \tag{7}$$

Let

$$M=\max_{1\leq j\leq n}|S_j|.$$

There are two cases.

Case 1: growth persists

Suppose (7) holds for every $k=1,\ldots,n-1$. The iterated part of the geometric lemma gives

$$|A_{n-1}|>\cos\alpha\,W_{n-1}. \tag{8}$$

On the other hand, (5) and the triangle inequality give

$$\begin{aligned} M W_{n-1} &\geq|S_n+b_1S_{n-1}+\cdots+b_{n-1}S_1|\\ &=|A_{n-1}|. \end{aligned}$$

Since $W_{n-1}>0$, comparison with (8) yields

$$M>\cos\alpha. \tag{9}$$

Case 2: first failure of growth

Otherwise let $k_0$ be the least index for which (7) fails. If $k_0>1$, then (7) held through $k_0-1$, so the iterated estimate gives

$$|A_{k_0-1}|>\cos\alpha\,W_{k_0-1}.$$

The same inequality also holds when $k_0=1$, because then its two sides are $1$ and $\cos\alpha$. Since (7) fails at $k_0$, the dichotomy forces (6) there. Using (4), followed by (6), gives

$$\begin{aligned} M W_{k_0-1} &\geq|S_{k_0}+b_1S_{k_0-1}+\cdots+b_{k_0-1}S_1|\\ &=|A_{k_0-1}-k_0b_{k_0}|\\ &\geq\sin\alpha\,|A_{k_0-1}|\\ &>\sin\alpha\cos\alpha\,W_{k_0-1}. \end{aligned}$$

Again $W_{k_0-1}>0$, and hence

$$M>\sin\alpha\cos\alpha. \tag{10}$$

Because $0<\sin\alpha<1$, (9) also implies (10). Thus (10) holds in both cases. Taking $\alpha=\pi/4$ gives

$$M>\sin\frac\pi4\cos\frac\pi4=\frac12,$$

which proves (1).

Source and dependency scope

The theorem and proof are on printed pp. 210--211, physical PDF pp. 224--225, of the published volume scan. The definition of $S_j$ and $R_n$ is on printed p. 209, physical p. 223. Every local inequality in the proof, including the planar geometry, appears here or on the linked lemma page. Only the standard Newton--Girard interface (2)--(3) remains external. No assertion about the sharp constant or the best presently known constant is part of this theorem.

Read depth. Claims checked: the statement, the definitions on p. 209 and equations (1)--(2) of the print were read clause by clause on the printed pages. The proof above is the paper's argument restated here step by step, with the case $n=1$ and the degenerate cases added; it is a compilation, not an independent review.

The paper's [[analysis/biro_1994_problem_turan_concerning_sums_powers_complex/theorem_2|Theorem 2]] (p. 212) refines (1) for $n\geq2$ to $\max_{1\leq j\leq n}|S_j|>\frac12+\frac1{8n}+\frac3{64n^2}$, its case of one prescribed $1$.

Bears on. Problem 519; (1) proves its exact existence statement with $c=1/2$.