Complex Numbers

Conjugate

If $z = x + iy$ is a complex number, $\bar{z} = x - iy$

Properties

For any complex number $z, z_1, z_2$, we have:
$ \begin{aligned} \text{(i)} \quad & \overline{\overline{z}} = z \\ \text{(ii)} \quad & z + \overline{z} = 2 \operatorname{Re}(z), \quad z + \overline{z} = 0 \Leftrightarrow z \text{ is purely imaginary.} \\ \text{(iii)} \quad & z - \overline{z} = 2i \operatorname{Im}(z), \quad z - \overline{z} = 0 \Leftrightarrow z \text{ is purely real.} \\ \text{(iv)} \quad & \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \\ \text{(v)} \quad & \overline{z_1 - z_2} = \overline{z_1} - \overline{z_2} \\ \text{(vi)} \quad & \overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2} \\ \text{(vii)} \quad & \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}, \quad z_2 \neq 0 \\ \text{(viii)} \quad & z_1 \overline{z_2} + \overline{z_1} z_2 = 2 \operatorname{Re}(\overline{z_1} z_2) = 2 \operatorname{Re}(z_1 \overline{z_2}) \\ \text{(ix)} \quad & \overline{z^n} = (\overline{z})^n \\ \text{(x)} \quad & \text{If } z = f(z_1), \text{ then } \overline{z} = f(\overline{z_1}) \\ \text{(xi)} \quad & \text{If } z = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}, \text{ then } \overline{z} = \begin{vmatrix} \overline{a_1} & \overline{a_2} & \overline{a_3} \\ \overline{b_1} & \overline{b_2} & \overline{b_3} \\ \overline{c_1} & \overline{c_2} & \overline{c_3} \end{vmatrix} \\ \text{(xii)} \quad & z \cdot \overline{z} = \{\operatorname{Re}(z)\}^2 + \{\operatorname{Im}(z)\}^2 \end{aligned} $