Given complex numbers $a,b,c$, we have that $|az^2 + bz +c| \leq 1$ holds true for any complex number $z, |z| \leq 1$. Find the maximum value of $|bc|$.
Problem
Source:
Tags: inequalities, trigonometry, calculus, function, complex numbers, algebra unsolved, algebra
Zhero
04.07.2011 05:03
We claim that the upper bound is $\frac{3 \sqrt{3}}{16}$.
By the maximum modulus principle, the maximum of $|az^2 + bz + c|$ over the closed unit disc must occur on the boundary of the unit disc, i.e., when $|z| = 1$. Without loss of generality, we may assume the maximum of $|az^2 + bz + c|$ over $|z| = 1$ is 1, or else we can scale $a,b,c$ by some constant factor larger than 1 so that $\max_{|z| = 1} |az^2 + bz + c| = 1$ and $|bc|$ increases. Furthermore, we can assume that $|a + b + c| = 1$; if $z_0$ maximizes $|az^2 + bz + c|$, taking $a' = a z_0^2$, $b' = b z_0$, and $c' = c$ gives $|b'c'| = |bc|$, $a' + b' + c' = 1$, and $|a'z^2 + b'z + c'| = |a(z_0 z)^2 + b(z z_0) + c| \leq 1$. Finally, by multiplying $a,b,c$ by some $e^{i \theta}$, we can suppose also that $a+b+c=1$.
For $|z| = 1$, let
\[ P(z) \&= |az^2 + bz + c|^2 = (az^2 + bz + c) (\overline{az^2 + bz + c}) \&= (az^2 + bz + c) (\frac{\overline{a}}{z^2} + \frac{\overline{b}}{z} + \overline{c}), \]and
\[ Q(z) = z^2 P(z) - z^2 = (az^2 + bz + c)(\overline{c}z^2 + \overline{b} z + \overline{a}) - z^2.\]
Let $f(\theta) = P(e^{i \theta})$. Note that $f(\theta) = 1 \iff Q(e^{i \theta}) = 0$. Also, if $f(\theta) = 1$ for some $\theta$, $f'(\theta) = 0$, or else $f(\theta \pm \epsilon) > 1$ for some positive real $\epsilon$ and some choice of $\pm$. In particular, we have that $f'(0) = 0$.
Now, suppose that $Q(z) = 0$, where $z = e^{i \theta}$. It would follow that $f(\theta) = 1$, whence
\begin{align*}
f'(\theta) = 0 &\implies i e^{i \theta} P'(e^{i \theta}) = 0 \implies P'(z) = 0 \\
&\implies \frac{(Q'(z) + 2z)z^2 - (Q(z) + z^2)2z}{z^4} = 0 \implies (Q'(z) + 2z)z^2 = 2z^3 \implies Q'(z) = 0,\end{align*}i.e., $Q$ must have a double root at $z$.
Since $Q(1) = 0$, $Q$ must have a double root at 1. We first calculate
\[ Q'(z) = (2a+b)(\overline{c}z^2 + \overline{b}z + \overline{a}) + (2\overline{c} + \overline{b})(az^2 + bz + c) - 2z, \]so
\begin{align*}
Q'(1) &= (2a + b)\overline{(a+b+c)} + \overline{(2c+b)}(a+b+c) - 2 \\
&= (2(1-b-c) + b) + \overline{2c+b} - 2 = 2(\overline{c} - c) + (\overline{b} - b) = 0.
\end{align*}
Thus, $2 \mbox{ Im } c = - \mbox{Im } b.$ Let $b = p + 2qi$ for some reals $p,q$; we must have $c = r - qi$ for some real $r$, and $a = (1-p-r) - qi$, since $a+b+c=1$. Substituting this into $Q$, we have
\[ Q(z) = (z-1)^2 Q_0(z), \]where $Q_0(z) = \alpha z^2 + \beta z + \overline{\alpha}$, and
\[ \alpha = -pr + q^2 - r^2 + r + i(-pq - 2qr + q), \]and
\[ \beta = -p^2 + p - 2pr - 2q^2 - 2r^2 + 2r. \]
Let $z_1, z_2$ be the two roots of $Q_0$, and let $\Delta = \beta^2 - 4 \alpha \overline{\alpha} = \beta^2 - 4|\alpha|^2$ be the discriminant of $Q_0$. We have $z_1 = \frac{-\beta + \sqrt{\Delta}}{2\alpha}$, and $z_2 = \frac{-\beta - \sqrt{\Delta}}{2 \alpha}$. Observing that $\beta$ and $\Delta$ are real, we note that if $\Delta < 0$,
\[ |z_1|^2 = |z_2|^2 = \frac{1}{4 |\alpha|^2} ( \beta^2 - \Delta^2 ) = 1, \]while $z_1 \neq z_2$. This contradicts the assumption that whenever $Q(z) = 0$, $z$ is a double root of $Q$ as well. Thus, we must have $\Delta \geq 0$.
We can compute $\Delta = (p-1)(p^3 + 4p^2 r - p^2 + 4pr^2 - 4pr + 4q^2) \geq 0$. Since $|az^2+bz+c| \leq 1$ at $z=-1$, we have $|a-b+c| = |1-2b| \leq 1$. But if $p > 1$ or $p < 0$, $|1 - 2b| > 1$, which is a contradiction, so $0 \leq p \leq 1$. Thus,
\[ p^3 + 4p^2 r - p^2 + 4pr^2 - 4pr + 4q^2 \leq 0. \]
We have $4|bc|^2 = 4|p + 2qi|^2 |r - qi|^2 = (p^2 + 4q^2)(4r^2 + 4q^2)$. Since $0 \leq 4q^2 \leq -p^3 - 4p^2 r + p^2 - 4pr^2 + 4pr$, we have
\[ 4|bc|^2 \leq (p^2 -p^3 - 4p^2 r + p^2 - 4pr^2 + 4pr)(4r^2 -p^3 - 4p^2 r + p^2 - 4pr^2 + 4pr) = p(1-p)(2-2r-p)(p+2r)^3. \]Note that $0 \leq p \leq 1$, so $2-2r-p \geq 0$. Thus, by AM-GM,
\[ \sqrt{p(1-p)} \leq \frac{p+1-p}{2} = \frac{1}{2} \implies p(1-p) \leq \frac{1}{4}, \]and
\[ \sqrt[4]{(2-2r-p) ( \frac{p + 2r}{3})^3} \leq \frac{2-2r-p + \frac{p+2r}{3} + \frac{p+2r}{3} + \frac{p+2r}{3}}{4} = \frac{1}{2} \implies (2-2r-p)(p+2r)^3 \leq \frac{27}{16}, \]so
\[ |bc|^2 \leq \frac{27}{256} \implies |bc| \leq \frac{3 \sqrt{3}}{16}. \]
We now show that when $p=r=\frac{1}{2}$ and $q = \frac{\sqrt{2}}{8}$, $a = (1-p-r) - qi = -\frac{\sqrt{2}}{8}i$, $b = p + 2qi = \frac{1}{2} + \frac{\sqrt{2}}{4}i$, and $c = r - qi = \frac{1}{2} - \frac{\sqrt{2}}{8}i$ satisfy $|az^2 + bz + c| \leq 1$ whenever $|z| = 1$. Because these are exactly the equality cases in all the inequalities above ($p =1-p$, $2-2r-p = \frac{p+2r}{3}$, and $q = \sqrt{-\frac{p^3 + 4p^2 r - p^2 + 4pr^2 - 4pr}{4}}$), we have $\Delta = 0$, so the two roots of $Q_0$ are $z_1 = z_2 = \frac{-1-2\sqrt{2}i}{3}$.
Let $\gamma = \arg z_1$. Recall that $f(\theta) = P(e^{i \theta})$. There are exactly two $\theta \in [0, 2\pi)$ with $f(\theta) = 1$: 0 and $\gamma$, and $f$ attains local extrema at both 0 and $\gamma$. We have $\pi < \gamma < \frac{3\pi}{2}$. Since $f(\pi), f(\frac{3\pi}{2}) \leq 1$, $f$ must attain local maxima at $1$ and $z_1$, i.e., $f(\theta) \leq 1$ for all $\theta$, so our proof is complete.
polya78
25.07.2011 00:01
I believe that solution is correct, and the detailed calculations are quite impressive. Here is a slightly quicker way to get the same result. Consider $\theta$ such that $\mid\theta\mid = 1$. Then $\mid az^2+bz+c\mid \le 1$ and $\mid a(z/\theta)^2+b(z/\theta)+c\mid\le1$, or $\mid az^2+b\theta z+c\theta^2\mid \le 1$. Let $v_1 = az^2+bz+c$ and $v_2= az^2+b\theta z+c\theta^2$. Then we have $\mid(v_1-v_2)\mid\le 2$, or $\mid bz(1-\theta)/2 + c(1-\theta^2)/2 \mid \le 1$. Now if $\mid b\prime z + c\prime\mid \le 1$, then by taking $z$ such that $b\prime z/c\prime$ is a positive real number, we must have that $\mid b\prime\mid +\mid c\prime\mid\le 1$, and so $\mid b\prime c\prime \mid \le 1/4$. Thus $\mid bc \mid * \mid(1-\theta)(1-\theta^2) \mid \le 1$. Let us try to maximize $ \mid(1-\theta)(1-\theta^2) \mid$. If $\theta = \alpha^2 =e^{2i\lambda}$, then $\mid(1-\theta)(1-\theta^2) \mid = \mid(1-\alpha^2)(1-\alpha^4) \mid = \mid (1/\alpha -\alpha)*(1/\alpha^2 - \alpha^2)\mid = 4 \mid (\sin \lambda)(\sin 2\lambda) \mid = 8 \mid (\cos\lambda)(1 - \cos^2 \lambda) \mid$. Using basic calculus, this has maximum value when $\cos\lambda = 1/\sqrt{3}$, which yields $\alpha = \frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{\sqrt{3}} i $, $\theta = -\frac{1}{3} + \frac{2\sqrt{2}}{3} i$, and $ \mid(1-\theta)(1-\theta^2) \mid = \frac{16}{3\sqrt{3}}$. So, $\mid bc \mid \le \frac{3\sqrt{3}}{16}$. It remains to prove that this maximum can be achieved. Note that if that is in fact true, then $b=1/(1-\theta)$, and $c=1/(1-\theta^2)$, and we must have $v_1=1$, or $a=1-1/(1-\theta)-1/(1-\theta^2)$. So we must prove that the function $f(z)= (1-1/(1-\theta)-1/(1-\theta^2)) *z^2 + 1/(1-\theta) *z + 1/(1-\theta^2)$ has modulus less than or equal to one if $\mid z \mid \le 1$. Unfortunately I can't find an elegant way of showing this, so a little brute force is in order. First note that it is enough to prove it if $\mid z \mid \ = 1$, and that it is also equivalent to proving that for $g(z)= (1-1/(1-\theta)-1/(1-\theta^2)) + 1/(1-\theta) *z + 1/(1-\theta^2) * z^2$. Plugging in $\theta =\alpha^2$, setting $y=z/\alpha$, and doing some calculations gets us to $g(z) = (i/4\sqrt{2})(3y^2+2\sqrt{3}y-1)$. If $y = e^{i\delta}$, then $\mid g(z) \mid ^2 = (1/32)\mid (22 + 8\sqrt{3} \cos \delta - 6\cos{2\delta} \mid = (1/32)\mid (28 + 8\sqrt{3} \cos \delta - 12\cos^2{\delta} \mid=(1/32)\mid (32 - 12(\cos\delta -1/\sqrt{3})^2) \mid\le 1$, and we are done.
iceillusion
27.05.2020 08:28
Let me give a solution with some geometric intuition: by the maximum modulus principle the assumption on $a,b,c$ is equivalent to \[\max_{\theta\in[0,2\pi)}|ae^{i\theta}+ce^{-i\theta}+b|\leq 1.\]The locus of $ae^{i\theta}+ce^{-i\theta}+b$ is an ellipse centred at $b$ with semi-major and semi-minor axes of length $|a|+|c|$ and $||a|-|c||$ respectively, and the assumption requires it to be within the unit disc. Therefore the problem is essentially about maximising the product of the centre's modulus and the average length of semi-major and semi-minor axes for ellipses within the unit disc. For any such ellipse we may try to (strictly) enlarge it with affine transformations fixing its centre. When this is impossible, by Bézout's theorem in algebraic geometry the ellipse is either tangent to the circle at $2$ different points or at a single point with intersection multiplicity in algebraic geometry $=4$. In the first case WLOG assume the tangent points have equal $y$-coordinate $v\in [0,1),$ then the ellipse has equation \[u(x^2+y^2-1)+(y-v)^2=0\]with $u>0$. In the second case we may assume the tangent point is $(0,1),$ then the ellipse has equation \[u(x^2+y^2-1)+(y-1)^2=0\]for some $u>0$ (in algebro-geometric terms the linear system of quadratic curves satisfying the intersection multiplicity condition has dimension 1 and contains $(y-1)^2$, so every curve in it should be given by such a linear combination), and we also denote $v=1$ in this case. Then the ellipse has centre $(0,\frac{v}{u+1})$ and semi axes length $\sqrt{\frac{u+1-v^2}{u+1}},$ $\frac{\sqrt{u(u+1-v^2)}}{u+1}$. By AM-GM we have \[\frac{v}{2(u+1)}(\sqrt{\frac{u+1-v^2}{u+1}}+\frac{\sqrt{u(u+1-v^2)}}{u+1})\leq\frac{\sqrt{u+1}+\sqrt{u}}{4(u+1)}\]with equality holds when $v=\sqrt{\frac{u+1}{2}}$, and by taking derivative we see when $u=\frac{1}{3}$ the RHS attains maximum $\frac{3\sqrt{3}}{16}$. Tracing back we get e.g. \[a=\frac{\sqrt{2}}{8},b=\frac{\sqrt{6}}{4}i,c=\frac{3\sqrt{2}}{8}\]satisfy the assumption and $|bc|=\frac{3\sqrt{3}}{16}$.