The Absolute Value/Modulus of a Complex Number

Consider a real number $x \in \mathbb{R}$. We know that the absolute value of $x$ is defined as follows:

(1)

\begin{align} \quad \mid x \mid = \left\{\begin{matrix} x & \mathrm{if} \: x \geq 0 \\ -x & \mathrm{if} \: x < 0 \end{matrix}\right. \end{align}

Geometrically, the absolute value of a real number denotes the distance $x$ is from the origin of the real number line. An analogous notion can be defined for the absolute value of a complex number.

Definition: Let $z = a + bi \in \mathbb{C}$. The Absolute Value or Modulus of $z$ is defined as $\mid z \mid = \sqrt{a^2 + b^2}$.

Recall that every complex number $z = a + bi \in \mathbb{C}$ can be regarded as a position vector (with initial point at the origin and terminating point at $(a, b)$) in the complex plane, and the length of that vector is given precisely as $|z| = \sqrt{a^2 + b^2}$ which can easily be worked out using the Pythagorean theorem as illustrated below:

For a worked out example, let $z = 5 + 12i$. Then the absolute value of $z$ is:

(2)

\begin{align} \quad \mid z \mid = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \end{align}

We will now state and prove some important properties regarding the absolute value complex numbers.

Proposition 1: Let $z = a + bi, w = c + di \in \mathbb{C}$. Then: a) $\mid z \mid = \mid \overline{z} \mid$. b) $-\mid z \mid \leq \mathrm{Re} (z) \leq \mid z \mid$. c) $-\mid z \mid \leq \mathrm{Im} (z) \leq \mid z \mid$.

• Proof: Let $z = a + bi \in \mathbf{C}$.

• Proof of a)

(3)

\begin{align} \quad \mid z \mid = \mid a + bi \mid = \sqrt{a^2 + b^2} = \sqrt{a^2 + (-b)^2} = \mid a - bi \mid = \mid \overline{z} \mid \quad \blacksquare \end{align}

• Proof of b) We have that:

(4)

\begin{align} \quad \mathrm{Re} (z) = a \leq \mid a \mid = \sqrt{a^2} \leq \sqrt{a^2 + b^2} = \mid z \mid \end{align}

• And similarly:

(5)

\begin{align} \quad \mathrm{Re} (z) = a \geq - \mid a \mid = -\sqrt{a^2} \geq -\sqrt{a^2 + b^2} = -\mid z \mid \end{align}

• Therefore $-\mid z \mid \leq \mathrm{Re} (z) \leq \mid z \mid$.

• Proof of c) We have that:

(6)

\begin{align} \quad \mathrm{Im} (z) = b \leq \mid b \mid = \sqrt{b^2} \leq \sqrt{a^2 + b^2} = \mid z \mid \end{align}

• And similarly:

(7)

\begin{align} \quad \mathrm{Im} (z) = b \geq - \mid b \mid = -\sqrt{b^2} \geq -\sqrt{a^2 + b^2} = -\mid z \mid \end{align}

• Therefore $-\mid z \mid \leq \mathrm{Im} (z) \leq \mid z \mid$.