Bound on complex triples satisfying some simple equations.

Question

Let $a,b,c \in \mathbb{C}$ be three complex numbers satifying the following equations : $$\left\{ \begin{array}{rl} a+b+c &= 3 \\ a^2+b^2+c^2 &=9 \end{array} \right.$$

I want to prove that $\max(|a|, |b|, |c|)\geq 2$.

I can formulate this as an optimisation problem for a multiple real variable function and use Lagrange multipliers, but it would probably be extremely long. The problem seems simple enough that there should be a direct and elegant proof.

So, any help would be very appreciated. Thanks.

Answer 1

We may assume $|a|\geq|b|\geq|c|$. We therefore introduce new variables $u$, $v\in{\mathbb C}$ such that $$b=a u, c= b v=a u v\qquad\bigl(|u|\leq1, \ |v|\leq 1)\ .$$ We then obtain $$3=a+b+c=a(1+u+uv)\tag{1}$$ and $$0=(a+b+c)^2-(a^2+b^2+c^2)=2(ab+ac+bc)=2a^2 u(1+v+uv)\ .$$ The case $u=0$ leads to $a=3>2$. Otherwise we have $1+v+uv=0$, or $$v=-{1\over 1+u}\ .\tag{2}$$ Since we want $|v|\leq1$ we need $|u+1|\geq1$. It follows that the admissible values for $u$ are in the closed unit disc from which the unit disc centered at $-1$ has been removed. Let $\bar \Omega$ be this moonshaped region. Given an $u\in\bar\Omega$ we have to calculate $v$ from $(2)$ and then have to verify in $(1)$ that $|1+u+uv|\leq{3\over2}$. This would imply that $|a|\geq2$.

Now $$1+u+uv={1+u+u^2\over1+u}=:f(u)\ .$$ In order to show that $|f(u)|\leq{3\over2}$ in $\bar\Omega$ it is sufficient to verify this inequality along the two boundary arcs. Along the arc $$\gamma_1:\quad t\mapsto u(t):=-1+e^{it}\qquad\left(-{\pi\over3}\leq t\leq{\pi\over3}\right)$$ we have $|1+u(t)|=1$, so that it remains to verify that $$\bigl|1+u(t)+u^2(t)\bigr|=2\cos t-1\in[0,1]\qquad\left(-{\pi\over3}\leq t\leq{\pi\over3}\right)\ .$$ Along the arc $$\gamma_2:\quad t\mapsto u(t):=e^{it}\qquad\left(-{2\pi\over3}\leq t\leq{2\pi\over3}\right)$$ we have to study the full complex $$w(t):=f\bigl(u(t)\bigr)={1+e^{it}+e^{2it}\over1+e^{it}}\qquad\left(-{2\pi\over3}\leq t\leq{2\pi\over3}\right)\ .$$ The following figure shows the Mathematica plot of this curve $t\mapsto w(t)$. We see that there is indeed a single point with $|w(t)|={3\over2}$, namely the point $w(0)={3\over2}$. It corresponds to $u=1$, $v=-{1\over2}$, and leads to the triple $(a,b,c):=(2,2,-1)$.

Answer 2

Plotting the Regions

Since $ab+bc+ca=\frac12\left((a+b+c)^2-\left(a^2+b^2+c^2\right)\right)=0$, Newton's identities say that $a$, $b$, and $c$ are roots of $$\newcommand{\Re}{\operatorname{Re}} z^3-3z^2-abc=0\tag1 $$ Suppose that $|a|\le|b|\le|c|\lt2$ $$ \begin{align} |c^3-3c^2| &=|c|^2|c-3|\tag{2a}\\ &=|abc|\tag{2b}\\ &\le|c|^3\tag{2c} \end{align} $$ That is, $|c-3|\le|c|$, which means that $c$ is closer to $3$ than it is to $0$. The dividing line between "closer to $0$" and "closer to $3$" is $\Re(z)=\frac32$. Therefore, in addition to $|c|\lt2$, we know that $\Re(c)\ge\frac32$.

Given $c$, we can use $(1)$ and Newton's identities to show that $a+b=3-c$ and $ab=c^2-3c$. Then we can then compute $$ \begin{align} a&=\frac{3-c-\sqrt{9+6c-3c^2}}2\tag{3a}\\ b&=\frac{3-c+\sqrt{9+6c-3c^2}}2\tag{3b} \end{align} $$ These could be switched, but it will be seen that, given $(3)$ and the possible locus of $c$, $\Re(b)\ge2$.

Here is a diagram of the regions where $c$, $a+b$, $ab$, $b$, and $a$ lie.

The yellow moon-shaped region, where $b$ lies, belies the assumption that $|a|\le|b|\le|c|\lt2$. Let us now prove that $\Re(b)\ge2$.

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Getting Real

Since $$ \begin{align} \cos(\theta) &=2\cos^2(\theta/2)-1\tag{4a}\\ &\le\cos^2(\theta/2)\tag{4b} \end{align} $$ if we use the square root with the branch cut along the negative real axis, where $\cos(\theta/2)\gt0$, we have $$ \overbrace{\Re\left(\sqrt{z}\right)}^{\sqrt{r}\cos(\theta/2)}\ge\overbrace{\sqrt{\Re(z)_+}}^{\sqrt{r}\sqrt{\cos(\theta)_+}}\tag5 $$ where $x_+=\max(x,0)$ for $x\in\mathbb{R}$.

With $c=x+iy$, $$ \Re(9+6c-3c^2)=9+6x-3x^2+3y^2\tag6 $$ Therefore, for $\frac32\le x\lt2$ $$ \begin{align} \Re(b) &=\Re\left(\frac{3-c+\sqrt{9+6c-3c^2}}2\right)\tag{7a}\\ &\ge\frac{3-x+\sqrt{9+6x-3x^2}}2\tag{7b}\\[12pt] &\ge2\tag{7c} \end{align} $$ Explanation:
$\text{(7a)}$: derived in $(3b)$
$\text{(7b)}$: apply $(5)$ and $(6)$
$\text{(7c)}$: the function is decreasing on $[1,3]$ and equals $2$ at $x=2$

$\text{(7c)}$ is a contradiction since we assumed $|a|\le|b|\le|c|\lt2$.