A calculus of the absurd
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14.1.3 The modulus of a complex number
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
\(\seteqnumber{0}{14.}{10}\)
\begin{equation}
\abs { \begin{pmatrix} x \\ y \end {pmatrix} } = \sqrt []{x^2 + y^2}
\end{equation}
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Theorem 14.1.2 For all \(z \in \mathbb {C}\), we have
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\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
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\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.
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Theorem 14.1.3 Let \(z, w \in \mathbb {C}\), then
\(\seteqnumber{0}{14.}{15}\)
\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),
\(\seteqnumber{0}{14.}{16}\)
\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
\(\seteqnumber{0}{14.}{19}\)
\begin{equation}
ac - bd - (ad + bc)i = \overline {zw}
\end{equation}
We can work backwards here, and obtain that
\(\seteqnumber{0}{14.}{20}\)
\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).