The conjugate of a sum of vectors in C^n is the sum of the conjugates

**Theorem:** For any \(\mathbf{z}, \mathbf{w} \in \mathbb{C}^n\), the conjugate of their sum equals the sum of their conjugates:

\[\overline{\mathbf{z} + \mathbf{w}} = \overline{\mathbf{z}} + \overline{\mathbf{w}}\]

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

\[\overline{(\mathbf{z} + \mathbf{w})}_j = \overline{z_j + w_j} = \overline{z_j} + \overline{w_j} = (\overline{\mathbf{z}} + \overline{\mathbf{w}})_j\]

The second equality uses the fact that conjugation distributes over addition for complex numbers: \(\overline{a + bi + c + di} = \overline{(a+c) + (b+d)i} = (a+c) - (b+d)i = (a - bi) + (c - di) = \overline{a+bi} + \overline{c+di}\).

**Example:**
- \(\mathbf{z} = (1+i, 2)\), \(\mathbf{w} = (3, -i)\)
- \(\mathbf{z} + \mathbf{w} = (4+i, 2-i)\) → \(\overline{\mathbf{z} + \mathbf{w}} = (4-i, 2+i)\)
- \(\overline{\mathbf{z}} + \overline{\mathbf{w}} = (1-i, 2) + (3, i) = (4-i, 2+i)\) ✓

**Related property:** Similarly, \(\overline{a \cdot \mathbf{z}} = \overline{a} \cdot \overline{\mathbf{z}}\) for scalar multiplication.Copy