The Set of Complex Numbers is a Field Examples 1

The Set of Complex Numbers is a Field Examples 1

Recall from The Set of Complex Numbers is a Field page that the set of complex numbers $\mathbb{C}$ with the operations of addition $+$ and multiplication $\cdot$ forms a field. We will now prove some basic results of this field using the field axioms.

Example 1

Prove that for every $z \in \mathbb{C}$ with $z \neq 0$ that the multiplicative inverse $z^{-1}$ of $z$ is unique.

Suppose that $z^{-1}$ and $z^{-1*}$ are both multiplicative inverses to $z \in \mathbb{C}$, $z \neq 0$. Then:

(1)
\begin{align} \quad z \cdot z^{-1} = 1 = z \cdot z^{-1*} \end{align}

Multiplying on the left by $z^{-1}$ gives us that $z^{-1} = z^{-1*}$, so the multiplicative inverse of $z$ is unique.

Example 2

Using the field axioms, prove that for all $z, w \in \mathbb{C}$ with $z, w \neq 0$ that $\displaystyle{\frac{1}{z \cdot w} = \frac{1}{z} \cdot \frac{1}{w}}$.

We equivalently want to prove that:

(2)
\begin{align} \quad (z \cdot w)^{-1} = z^{-1} \cdot w^{-1} \end{align}

Note that:

(3)
\begin{align} \quad (z^{-1} \cdot w^{-1}) \cdot (z \cdot w) &= (z^{-1} \cdot w^{-1} ) \cdot (w \cdot z) \quad (\mathrm{Multiplicative \: Commutativity}) \\ &= z^{-1} \cdot ( w^{-1} \cdot w) \cdot z \quad (\mathrm{Multiplication \: Associativity}) \\ &= z^{-1} \cdot 1 \cdot z \quad (\mathrm{Multiplicative \: Inverses}) \\ &= z^{-1} \cdot z \quad (\mathrm{Multiplicative \: Identity}) \\ &= 1 \quad (\mathrm{Multiplicative \: Inverses}) \end{align}

So $(z \cdot w)$ is the multiplicative inverse of $(z^{-1} \cdot w^{-1})$ and conversely, $z^{-1} \cdot w^{-1}$ is the multiplicative inverse of $z \cdot w$. But $(z \cdot w)^{-1}$ is also the multiplicative inverse of $z \cdot w$ and so by the uniqueness of multiplicative inverses we see that:

(4)
\begin{align} \quad (z \cdot w)^{-1} = z^{-1} \cdot w^{-1} \end{align}

Example 3

Suppose that $\displaystyle{\frac{x - yi}{x + yi} = a + bi}$. Prove that then $a^2 + b^2 = 1$.

Let $z = x + yi$. Then $\overline{z} = x - yi$ and so:

(5)
\begin{align} \quad \frac{\overline{z}}{z} = a + bi \end{align}

Take the modulus of both sides of the equation above to get:

(6)
\begin{align} \quad \biggr \lvert \frac{\overline{z}}{z} \biggr \rvert & = \mid a + bi \mid \\ \quad \frac{\mid \overline{z} \mid}{\mid z \mid} & = \sqrt{a^2 + b^2} \end{align}

We know that $\mid z \mid = \mid \overline{z} \mid$ for all $z \in \mathbb{C}$, and so $\displaystyle{\frac{\mid \overline{z} \mid}{\mid z \mid} = 1}$ (provided the denominator is nonzero, and so:

(7)
\begin{align} \quad 1 & = \sqrt{a^2 + b^2} \\ \quad 1 & = a^2 + b^2 \end{align}