Problem 3. Let $\alpha, \beta \in \mathbb{R}$. Determine the maximum value of the expression

$$
|\alpha x+\beta y|+|\alpha x-\beta y|
$$

in each of the following cases:
a) $x, y \in \mathbb{R}$, such that $|x| \leq 1$ and $|y| \leq 1$;
b) $x, y \in \mathbb{C}$, such that $|x| \leq 1$ and $|y| \leq 1$.
Solution: a) Since for any real numbers $u, v$ we have $|u+v|+|u-v| \in \{ \pm 2 u, \pm 2 v\}$, we deduce that $|\alpha x+\beta y|+|\alpha x-\beta y| \leq \max \{2|\alpha|, 2|\beta|\} \ldots \ldots 2 \mathbf{p}$

This is the maximum value, since equality is obtained, for example, when $x=y=1$.

b) We have the equality $|\alpha x+\beta y|^{2}+|\alpha x-\beta y|^{2}=2|\alpha x|^{2}+2|\beta y|^{2}$. But

$(|\alpha x+\beta y|+|\alpha x-\beta y|)^{2} \leq 2\left(|\alpha x+\beta y|^{2}+|\alpha x-\beta y|^{2}\right)$, from which $|\alpha x+\beta y|+$

![](https://cdn.mathpix.com/cropped/2024_06_07_b5f4bd8b782e050f8d06g-3.jpg?height=57&width=1211&top_left_y=980&top_left_x=457)

The maximum value $2 \sqrt{\alpha^{2}+\beta^{2}}$ can be obtained for $x=1$ and $y=i . \ldots \mathbf{1 p}$