**Theorem:** For any scalar \(a \in \mathbb{C}\) and any vector \(\mathbf{z} \in \mathbb{C}^n\), the conjugate of their product equals the product of their conjugates:

\[\overline{a \cdot \mathbf{z}} = \overline{a} \cdot \overline{\mathbf{z}}\]

**Proof:** For each component \(j\):

\[\overline{(a \cdot \mathbf{z})}_j = \overline{a \cdot z_j} = \overline{a} \cdot \overline{z_j} = (\overline{a} \cdot \overline{\mathbf{z}})_j\]

The second equality uses the fact that conjugation distributes over multiplication for complex numbers: \(\overline{(a+bi)(c+di)} = \overline{(ac-bd) + (ad+bc)i} = (ac-bd) - (ad+bc)i = (a-bi)(c-di) = \overline{a+bi} \cdot \overline{c+di}\).

**Example:**
- \(a = 1+i\), \(\mathbf{z} = (2, i)\)
- \(a \cdot \mathbf{z} = (2+2i, -1+i)\) → \(\overline{a \cdot \mathbf{z}} = (2-2i, -1-i)\)
- \(\overline{a} \cdot \overline{\mathbf{z}} = (1-i) \cdot (2, -i) = (2-2i, -1-i)\) ✓

**Special case:** When \(a \in \mathbb{R}\), we have \(\overline{a} = a\), so \(\overline{a \cdot \mathbf{z}} = a \cdot \overline{\mathbf{z}}\).