# 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)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)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)**

**Proof of b)**We have that:

- And similarly:

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

**Proof of c)**We have that:

- And similarly:

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