Sum of Complex Conjugates
From ProofWiki
Theorem
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\overline {z}$ be the complex conjugate of the complex number $z$.
Then:
- $\overline {z_1 + z_2} = \overline {z_1} + \overline {z_2}$
Proof
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \overline {z_1 + z_2}\) | \(=\) | \(\displaystyle \overline {\left({x_1 + x_2}\right) + i \left({y_1 + y_2}\right)}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x_1 + x_2}\right) - i \left({y_1 + y_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({x_1 - i y_1}\right) + \left({x_2 - i y_2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \overline {z_1} + \overline {z_2}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$