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

    \begin{equation} |z| = \sqrt []{x^2 + y^2} \end{equation}

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

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

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

    \begin{equation} z \overline {z} = \abs {z}^2. \end{equation}

    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 5.2) 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 5.2),

\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

\begin{equation} ac - bd - (ad + bc)i = \overline {zw} \end{equation}

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 5.2 for more on this).