Fill in conjugate-of-scalar-mult theorem

# The conjugate of vector-scalar multiplication in C^n is the product of the conjugates.

Put content here**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:
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\[\overline{a \cdot \mathbf{z}} = \overline{a} \cdot \overline{\mathbf{z}}\]
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**Proof:** For each component \(j\):
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\[\overline{(a \cdot \mathbf{z})}_j = \overline{a \cdot z_j} = \overline{a} \cdot \overline{z_j} = (\overline{a} \cdot \overline{\mathbf{z}})_j\]
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The second equality uses the fact that conjugation distributes over multiplication for complex numbers: \(\overline{(a+bi)(c+di)} = \overline{(ac-bd) + (ad+bc)i} = (ac-bd) - (ad+bc)i = (a-bi)(c-di) = \overline{a+bi} \cdot \overline{c+di}\).
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**Example:**
- \(a = 1+i\), \(\mathbf{z} = (2, i)\)
- \(a \cdot \mathbf{z} = (2+2i, -1+i)\) → \(\overline{a \cdot \mathbf{z}} = (2-2i, -1-i)\)
- \(\overline{a} \cdot \overline{\mathbf{z}} = (1-i) \cdot (2, -i) = (2-2i, -1-i)\) ✓
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**Special case:** When \(a \in \mathbb{R}\), we have \(\overline{a} = a\), so \(\overline{a \cdot \mathbf{z}} = a \cdot \overline{\mathbf{z}}\).

# Parents

* Algebraic properties of R^n (or C^n)