\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let \(f(z) = z^3 + az^2 + bz + c\). If we restrict the domain to the reals, we have \[ \lim _{x \to \infty } f(x) = \infty , \lim _{x \to -\infty } f(x) = -\infty . \]
By the definition of a limit, this means that \(f(x) > 0\) for sufficiently big \(x\)s (say, for all \(x \geq A\)), and \(f(x) < 0\) for sufficiently small \(x\)s (say, for all \(x \leq B\)).
Since \(f\) is continuous on \([B, A] \subset \RR \), and \(f(B) < 0\), \(f(A) > 0\). This means that for some \(\xi \in (B, A) \subset \RR \) such that \(f(\xi ) = 0\), which finishes our proof.
By Vieta’s Theorem, we have \begin {align*} z_1 + z_2 + z_3 & = -a, \\ z_1 z_2 + z_1 z_3 + z_2 z_3 & = b, \\ z_1 z_2 z_3 & = -c. \end {align*}
Therefore, we have \(S_1 = -a\) and \(a = -S_1\). Notice that \begin {align*} \frac {S_1^2 - S_2}{2} & = \frac {(z_1 + z_2 + z_3)^2 - (z_1^2 + z_2^2 + z_3^2)}{2} \\ & = \frac {2 \cdot (z_1 z_2 + z_1 z_3 + z_2 z_3)}{2} \\ & = z_1 z_2 + z_1 z_3 + z_2 z_3 \\ & = b. \end {align*}
This means \begin {align*} a & = - S_1, \\ b & = \frac {S_1^2 - S_2}{2}. \end {align*}
Also, notice that \begin {align*} -S_1^3 + 3 S_1 S_2 - 2 S_3 & = - (z_1 + z_2 + z_3)^3 + 3 (z_1 + z_2 + z_3) (z_1^2 + z_2^2 + z_3^2) - 2(z_1^3 + z_2^3 + z_3^3) \\ & = -(z_1^3 + z_2^3 + z_3^3 + 3z_1 z_2^2 + 3z_1 z_3^2 + 3 z_2 z_1^2 + 3 z_2 z_3^2 + 3 z_3 z_1^2 + 3 z_3 z_2^2 + 6 z_1 z_2 z_3) \\ & \phantom {=} + 3(z_1^3 + z_2^3 + z_3^3 + z_1 z_2^2 + z_1 z_3^2 + z_2 z_1^2 + z_2 z_3^2 + z_3 z_1^2 + z_3 z_2^2) \\ & \phantom {=} - 2(z_1^3 + z_2^3 + z_3^3) \\ & = -6 z_1 z_2 z_3 \\ & = 6c, \end {align*}
as desired.
Consider the complex numbers \(z_k = r_k \exp (i \theta _k)\) for \(k = 1, 2, 3\).
This means that \(z_k^n = r_k^n \exp (i n \theta _k)\) by de Moivre’s theorem, hence \[ \im z_k^n = r_k^n \sin (n \theta _k). \]
This converts our condition to \begin {align*} \im \sum _{k = 1}^{3} z_k^n = 0 \end {align*}
for \(n = 1, 2, 3\).
Therefore, \(S_1, S_2, S_3\) are real, and therefore, so are \(a, b, c\).
Hence, by part (i), there must be some \(k\) such that \(z_k\) is real, which means \(\theta _k\) is some multiple of \(2\pi \).
Since \(\theta _k \in (-\pi , \pi )\), we must have \(\theta _k = 0\) for such.
If \(\theta _1 = 0\), \(z_1 \in \RR \). This therefore means that \(z_1^n \in \RR \), and hence \[ \im \sum _{k = 2}^{3} z_k^n = 0 \] for \(n = 1, 2, 3\).
Consider the polynomial \((z - z_2)(z - z_3) = 0\), and let the expansion be \(z^2 + pz + q = 0\).
By Vieta’s Theorem, we have \begin {align*} z_2 + z_3 & = -p, \\ z_2 z_3 & = q. \end {align*}
This therefore means that \begin {align*} p & = - (z_2 + z_3), \\ 2q & = (z_2 + z_3)^2 - (z_2^2 + z_3^2). \end {align*}
If \(z_2 + z_3 \in \RR \) and \(z_2^2 + z_3^2 \in \RR \), then \(p, q \in \RR \), and \(z_2, z_3\) are solutions to a real quadratic (polynomial).
Hence, the first case is \(z_2, z_3\) are both real, which gives \(\theta _2 = \theta _3 = 0\) since \(r_k > 0\), and hence \(\theta _2 = -\theta _3\).
The other case where \(z_2, z_3\) are complex congruent to each other gives \(\theta _2 = - \theta _3 + 2k\pi \) where \(k \in \ZZ \) due to \(r_k > 0\). But since \(\theta _2, \theta _3 \in (-\pi , \pi )\), it must be the case that \(\theta _2 = -\theta _3\), since the width of the interval is exactly \(2\pi \), and it is an open interval.
This finishes our proof.