Conjugation of Complex Numbers

# Conjugation of Complex Numbers

Recall from the Addition and Multiplication of Complex Numbers page that if $z = a + bi, w = c + di \in \mathbb{C}$ then the sum $z + w$ by addition is defined as:

(1)
\begin{align} \quad z + w = (a + c) + (b + d)i \end{align}

Furthermore, the product $z \cdot w$ (or simply $zw$) by multiplication is defined as:

(2)
\begin{align} \quad zw = (ac - bd) + (ad + bc)i \end{align}

We will now look at a very simple operation known as complex conjugation (or simply conjugation) of a complex number.

 Definition: If $z = a + bi \in \mathbb{C}$ then the Conjugate of $z$ is the complex number $\overline{z} = a - bi$.

In other words, if $z$ is a complex number then the conjugate of $z$ is the complex number whose real part is $\mathrm{Re}(z)$ and whose imaginary part is $-\mathrm{Im}(z)$.

For example, if $z = 4 - 2i$ then:

(3)
\begin{align} \quad \overline{z} = 4 + 2i \end{align}
 Proposition 1 (Properties of Conjugation): Let $z = a + bi \in \mathbb{C}$. Then: a) $z \in \mathbb{R}$ if and only if $z = \overline{z}$. b) $z + \overline{z} = 2a$. c) $z - \overline{z} = 2bi$. d) $z \cdot \overline{z} = a^2 + b^2$.
• Proof of a) $\Rightarrow$ Suppose that $z \in \mathbb{R}$. Then $\mathrm{Im}(z) = b = 0$, so:
(4)
\begin{align} \quad z = a + bi = a + 0i = a - 0i = \overline{z} \end{align}
• $\Leftarrow$ Suppose that $z = \overline{z}$. Then:
(5)
\begin{align} \quad a + bi &= a - bi \\ \quad bi &= -bi \\ \quad 2bi &= 0 \\ \quad bi &= 0 \end{align}
• Since $i \neq 0$, this implies that $\mathrm{Im}(z) = b = 0$, so $z \in \mathbb{R}$. $\blacksquare$
• Proof of b) Let $z = a + bi$. Then:
(6)
\begin{align} \quad z + \overline{z} = (a + bi) + (a - bi) = (a + a) + (b - b)i = 2a + 0i = 2a \quad \blacksquare \end{align}
• Proof of c) Let $z = a + bi$. Then:
(7)
\begin{align} \quad z - \overline{z} = (a + bi) - (a - bi) = (a - a) + (b - (-b))i = 2bi \quad \blacksquare \end{align}
• Proof of d) Let $z = a + bi$. Then:
(8)
\begin{align} \quad z \cdot \overline{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = (a^2 + b^2) + (ab - ab)i = a^2 + b^2 \quad \blacksquare \end{align}