0
$\begingroup$

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 $y \in \mathbb{R}$.

$\endgroup$
2
$\begingroup$

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: enter image description here

$\endgroup$
  • 1
    $\begingroup$ Are those all solutions to $a\bar a+b\bar b=1$? $a=e^{i\alpha}\cos\psi,b=e^{i\beta}\sin\psi$ is sufficient. Is it necessary? $\endgroup$ – marty cohen Oct 8 at 18:58
  • 1
    $\begingroup$ @martycohen: From $|a|^2+|b|^2=1$ it follows that we can write $|a|=\cos\psi$, $|b|=\sin\psi$ with some $\psi\in[0,\pi/2]$. $\endgroup$ – Christian Blatter Oct 8 at 19:04
  • $\begingroup$ It's interesting that calculus is not needed to determine the max. $\endgroup$ – marty cohen Oct 8 at 19:28
  • $\begingroup$ Why is $\mathcal{Q}$ maximal when $\beta - \alpha = \xi - \eta \mod 2\pi$? $\endgroup$ – Viktor Glombik Oct 8 at 20:00
  • $\begingroup$ Because the two summands then show in the same direction. – Please restrict edits of other people's answers to the correction of typos, and similar. $\endgroup$ – Christian Blatter Oct 9 at 7:03
0
$\begingroup$

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) $.

$\endgroup$
  • $\begingroup$ Aren't you missing some absolute values around the $y \in \mathbb{C}$? $\endgroup$ – Viktor Glombik Oct 8 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.