Find $\sup_{a a^* + b b^* = 1} | a^2x + aby |$ for fixed $x,y \in \mathbb{C}$, $|x| = 1$, $y \ne 0$.

Question

While trying to come up with an answer to this question without using the theorem used in the existing answer, whose proof is non-trivial, I tried to find $$ \tag{1} \sup_{a a^* + b b^* = 1} | a^2x + aby | $$ for some fixed complex numbers $x, y \in \mathbb{C}$ In the question the conditions $|x| = 1$ and $y \ne 0$ are added but I am interested if there's also a more general solution without those constrains. The task of the question is to prove that $$(1)> |x|.$$

My progress Aiming to use Lagrange multipliers, I expanded the term $a^2x + aby$ in terms of $n_1 := \Re(n)$ and $n_2 := \Im(n)$ for $n \in \{a,b,x,y\}$ to then use that $$\tag{2} |z| = \sqrt{z_1^2 + z_2^2} $$ for all $z \in \mathbb{C}$. I ended up with $$ \Re(a^2x + aby) = a_1 b_1 y_1 + a_1 b_2 y_2 + a_2 b_1 y_2 - a_2 b_1 y_2 + a_2^1 x_1 + a_2^2 x_1 $$ and $$ \Im(a^2 x + aby) = a_1 b_1 y_2 - a_1 b_2 y_1 + a_2 b_1 y_1 + a_2 b_2 y_2 + a_1^2 x_2 + a_2^2 x_2. $$ Plugging this into (2) yields the square root of more than 40 terms. Is there a nice way to factor this or to approach this?

If this were only a real problem, we'd have $$ \sup_{a^2 + b^2 = 1} ax^2 + a b y = \sup_{|a| \le 1} a + a \sqrt{1 - a^2} y $$ Setting the derivative w.r.t to $a$ equal to zero we obtain $$ 0 = \sqrt{1 - a^2} y + 1 - \frac{a^2 y}{\sqrt{1 - a^2}}, $$ giving $$ a^2 = \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2} + \frac{1}{2}. $$ Plugging the positive root in gives $$ a + a \sqrt{1 - a^2} y = \left(\sqrt{\frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2} + \frac{1}{2}}\right)\left(1 + \sqrt{\frac{1}{2} - \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2}} \cdot y\right), $$ which looks approximately linear in $y$ when plotted with WolframAlpha. Unfortunately, $\Re\left(\sqrt{\frac{1}{2} - \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2}}\right) = 0$ for all $y \in \mathbb{R}$.

Answer 1

Write $$a=e^{i\alpha}\cos(\psi),\qquad b=e^{i\beta}\sin(\psi),\qquad 0\leq\psi\leq{\pi\over2}\ .$$ Then $a\bar a+b\bar b=1$. We have to find the sup of the quantity $$Q:=\left|e^{2i\alpha}\cos^2(\psi) \>x+e^{i(\alpha+\beta)}\cos(\psi)\sin(\psi)\>y\right|\ ,$$ whereby $\alpha$, $\beta$, and $\psi$ are variables. As $|x|=1$ and $y\ne0$ we can write $$x=e^{i\xi},\qquad y=|y|e^{i\eta}\ ,$$ so that $$ Q =\left|\cos^2(\psi) \>e^{i\xi}+\cos(\psi)\sin(\psi)\>|y|\>e^{i(\beta-\alpha+\eta)}\right|\ . $$ Since $\xi$ and $\eta$ are given, and $0\leq\psi\leq{\pi\over2}$, this $Q$ is maximal when $\beta-\alpha=\xi-\eta$ mod $2\pi$. It follows that $$\max_{\alpha, \beta,\psi}Q =\max_{0\leq\psi\leq\pi/2}\left(\cos^2(\psi)+\cos(\psi)\sin(\psi)\>|y|\right)\ .$$ In this way we are left with a single variable ($\psi$) extremal problem: We have $$ \frac{\textrm{d}}{\textrm{d} \phi} (\cos^2(\psi)+\cos(\psi)\sin(\psi) \cdot |y|) = | y | \cos(2\phi) - \sin(2\phi) \overset{!}{=} 0 \implies x = \frac{1}{2} \tan^{-1}(|y|) $$ (and $\frac{\textrm{d}^2}{\textrm{d} \phi^2} \big(\cos^2(\psi)+\cos(\psi)\sin(\psi)|y|\big) \big|_{x = \frac{1}{2} \tan^{-1}(|y|)} < 0$). Plugging in gives $$ \max_{\alpha, \beta, \phi} Q = \frac{\sqrt{|y|^2 + 1} + 1}{2}, $$ which is always greater as 1 and only equal to one if $y = 0$, which we have excluded. Plotted as a function of its real $(=x)$ and imaginary part $(=y)$, it looks like this:

Answer 2

Here is the solution to Christian Blatter's maximization problem, writing $p$ for $\psi$.

$f(p) =\cos^2(p)+\cos(p)\sin(p)y =\frac12(\cos(2p)+1)+\frac12\sin(2p)y $ so maximizing this is the same as $g(q) =\cos(q)+\sin(q)y $ with $p = q/2$.

Let $\tan(r) = y$, so $\sin(r) =\frac{y}{\sqrt{1+y^2}} $ and $\cos(r) =\frac{1}{\sqrt{1+y^2}} $.

Then

$\begin{array}\\ g(q) &=\cos(q)+\sin(q)y\\ &=\sqrt{1+y^2}(\frac1{\sqrt{1+y^2}}\cos(q)+\frac{y}{\sqrt{1+y^2}}\sin(q))\\ &=\sqrt{1+y^2}(\cos(r)\cos(q)+\sin(r)\sin(q))\\ &=\sqrt{1+y^2}\cos(r-q)\\ \end{array} $

This is maximized when $r-q=0$ so that $q = r =\arctan(y) $ or $2p = r $ or $p = \frac12\arctan(y) $.