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.