# A calculus of the absurd

##### 14.1.3 The modulus of a complex number
• Definition 14.1.1 We define the “modulus” of a complex number $$z = x + iy$$ as

$$|z| = \sqrt []{x^2 + y^2}$$

If we plot a complex number, then the modulus is the “length” of the complex number. This is exactly the same as how the magnitude of the vector in $$\mathbb {R}^2$$ (all real-number valued coordinates. e.g. $$(1, 2)$$, $$(\pi , 6)$$, etc.) is defined as

$$\abs { \begin{pmatrix} x \\ y \end {pmatrix} } = \sqrt []{x^2 + y^2}$$

• Theorem 14.1.2 For all $$z \in \mathbb {C}$$, we have

$$z \overline {z} = \abs {z}^2.$$

That is, the value of $$z$$ multiplied by its complex conjugate is the modulus of $$z$$.

Proof: we can prove this by direct computation. If $$z$$ is a complex number then there exist $$x$$ and $$y$$ such that $$z = x + iy$$, and therefore

\begin{align} z\overline {z} &= (x + iy) \times (x - iy) \\ &= x^2 - (iy)^2 \\ &= x^2 + y^2 \end{align}

Note that we used the difference of two squares identity (see Section 6.2) to show this.

• Theorem 14.1.3 Let $$z, w \in \mathbb {C}$$, then

\begin{align} \overline {zw} = \overline {z} \times \overline {w} \end{align}

Proof: let $$z = a + bi$$ and $$w = c + di$$, then we proceed with the standard method of proving identities (Section 6.2),

\begin{align} \overline {z} \overline {w} &= (a - bi)(c - di) \\ &= ac - (ad + bc)i - bd \\ &= ac - bd - (ad + bc)i \end{align}

What remains is to prove that

$$ac - bd - (ad + bc)i = \overline {zw}$$

We can work backwards here, and obtain that

\begin{align} \overline {zw} &= \overline {(a + bi)(c + di)} \\ &= \overline {ac + (ad + bc)i - bd} \\ &= \overline {ac - bd + (ad + bc)i} \\ &= ac - bd - (ad + bc)i. \end{align}

Which is what we needed to prove (note that it is of paramount importance that our proof - which it does - flows in both directions; that is, that every step is reversible, see Section 6.2 for more on this).