Let $ \alpha ,\beta $ be real numbers. Find the greatest value of the expression
$$ |\alpha x +\beta y| +|\alpha x-\beta y| $$
in each of the following cases:

[b]a)[/b] $ x,y\in \mathbb{R} $ and $ |x|,|y|\le 1 $
[b]b)[/b] $ x,y\in \mathbb{C} $ and $ |x|,|y|\le 1 $
### Part (a)
We need to find the greatest value of the expression \( |\alpha x + \beta y| + |\alpha x - \beta y| \) given that \( x, y \in \mathbb{R} \) and \( |x|, |y| \le 1 \).

1. **Using the Cauchy-Schwarz Inequality:**
   The Cauchy-Schwarz inequality states that for any vectors \( \mathbf{u} \) and \( \mathbf{v} \) in an inner product space,
   \[
   |\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\| \|\mathbf{v}\|
   \]
   Here, let \( \mathbf{u} = (\alpha, \beta) \) and \( \mathbf{v} = (x, y) \). Then,
   \[
   |\alpha x + \beta y| \leq \sqrt{\alpha^2 + \beta^2} \sqrt{x^2 + y^2}
   \]
   Since \( |x|, |y| \leq 1 \), we have \( x^2 + y^2 \leq 2 \). Therefore,
   \[
   |\alpha x + \beta y| \leq \sqrt{\alpha^2 + \beta^2} \sqrt{2}
   \]

2. **Applying the same for \( |\alpha x - \beta y| \):**
   Similarly,
   \[
   |\alpha x - \beta y| \leq \sqrt{\alpha^2 + \beta^2} \sqrt{2}
   \]

3. **Summing the inequalities:**
   \[
   |\alpha x + \beta y| + |\alpha x - \beta y| \leq \sqrt{2} \sqrt{\alpha^2 + \beta^2} + \sqrt{2} \sqrt{\alpha^2 + \beta^2} = 2\sqrt{2} \sqrt{\alpha^2 + \beta^2}
   \]

4. **Checking if the bound is achievable:**
   To check if this bound is achievable, consider \( x = \pm 1 \) and \( y = \pm 1 \). For example, if \( x = 1 \) and \( y = 1 \),
   \[
   |\alpha \cdot 1 + \beta \cdot 1| + |\alpha \cdot 1 - \beta \cdot 1| = |\alpha + \beta| + |\alpha - \beta|
   \]
   If \( \alpha \) and \( \beta \) are such that \( \alpha = \beta \), then
   \[
   |\alpha + \beta| + |\alpha - \beta| = |2\alpha| + |0| = 2|\alpha|
   \]
   If \( \alpha \) and \( \beta \) are such that \( \alpha = -\beta \), then
   \[
   |\alpha + \beta| + |\alpha - \beta| = |0| + |2\alpha| = 2|\alpha|
   \]
   Thus, the bound \( 2\sqrt{2} \sqrt{\alpha^2 + \beta^2} \) is achievable.

### Part (b)
We need to find the greatest value of the expression \( |\alpha x + \beta y| + |\alpha x - \beta y| \) given that \( x, y \in \mathbb{C} \) and \( |x|, |y| \le 1 \).

1. **Considering complex numbers:**
   For complex numbers \( x \) and \( y \), we have \( |x| \leq 1 \) and \( |y| \leq 1 \). The modulus of a complex number \( z \) is given by \( |z| = \sqrt{\text{Re}(z)^2 + \text{Im}(z)^2} \).

2. **Using the triangle inequality:**
   The triangle inequality for complex numbers states that for any complex numbers \( z_1 \) and \( z_2 \),
   \[
   |z_1 + z_2| \leq |z_1| + |z_2|
   \]
   Applying this to our expression, we get:
   \[
   |\alpha x + \beta y| + |\alpha x - \beta y| \leq |\alpha x| + |\beta y| + |\alpha x| + |\beta y| = 2|\alpha x| + 2|\beta y|
   \]

3. **Maximizing the expression:**
   Since \( |x| \leq 1 \) and \( |y| \leq 1 \), we have \( |\alpha x| \leq |\alpha| \) and \( |\beta y| \leq |\beta| \). Therefore,
   \[
   2|\alpha x| + 2|\beta y| \leq 2|\alpha| + 2|\beta|
   \]

4. **Checking if the bound is achievable:**
   To check if this bound is achievable, consider \( x = 1 \) and \( y = 1 \). Then,
   \[
   |\alpha x + \beta y| + |\alpha x - \beta y| = |\alpha + \beta| + |\alpha - \beta|
   \]
   If \( \alpha \) and \( \beta \) are such that \( \alpha = \beta \), then
   \[
   |\alpha + \beta| + |\alpha - \beta| = |2\alpha| + |0| = 2|\alpha|
   \]
   If \( \alpha \) and \( \beta \) are such that \( \alpha = -\beta \), then
   \[
   |\alpha + \beta| + |\alpha - \beta| = |0| + |2\alpha| = 2|\alpha|
   \]
   Thus, the bound \( 2|\alpha| + 2|\beta| \) is achievable.

The final answer is \( \boxed{ 2\sqrt{2} \sqrt{\alpha^2 + \beta^2} } \) for part (a) and \( 2|\alpha| + 2|\beta| \) for part (b).