The Conjugate of a Complex Number

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 $z \in \mathbb{C}$. Then: a) $z$ is a real number if and only if $z = \bar{z}$. b) $z$ is a purely imaginary number if and only if $z = -\bar{z}$. c) $z + \overline{z} = 2\mathrm{Re}(z)$. c) $z - \overline{z} = 2\mathrm{Im}(z)i$. d) $z \cdot \overline{z} = |z|^2$.

• Proof: Let $z = a + bi$.

• Proof of a) $\Rightarrow$ If $z$ is a real number then $b = 0$. So $z = a$. But then $\bar{z} = a - bi = a - 0i = a = z$. $\Leftarrow$ If $z = \bar{z}$ then $a + bi = a - bi$. So $2bi = 0$ implying that $b = 0$. So $z$ is a real number. $\blacksquare$

• Proof of b) $\Rightarrow$ If $z$ is a purely imaginary number then $a = 0$. So $z = bi$. But then $\bar{z} = a - bi = 0 -bi = -bi = -\bar{z}$. $\Leftarrow$ Suppose that $z = -\bar{z}$. Then $a + bi = -a + bi$. So $a = -a$ implying that $2a = 0$. So $a = 0$. Thus $z$ is a purely imaginary number. $\blacksquare$

• Proof of c) We have that:

(4)

\begin{align} \quad z + \bar{z} = (a + bi) + (a - bi) = 2a = 2\mathrm{Re}(z) \quad \blacksquare \end{align}

• Proof of d) We have that:

(5)

\begin{align} \quad z - \bar{z} = (a + bi) - (a - bi) = 2bi = 2\mathrm{Im}(z)i \quad \blacksquare \end{align}

• Proof of e) We have that:

(6)

\begin{align} \quad z\bar{z} = (a + bi)(a - bi) = (a^2 + b^2) + (-ab + ba)i = a^2 + b^2 = |z|^2 \quad \blacksquare \end{align}

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