Non-Zero Complex Numbers are Closed under Multiplication/Proof 3
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Theorem
The set of non-zero complex numbers is closed under multiplication.
Proof
Equivalently this is to say:
- $z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$
Let $z_1 z_2 = 0$.
\(\ds z_1\) | \(=\) | \(\ds \tuple {x_1, y_1}\) | Definition 2 of Complex Number: for some $x_1, y_1 \in \R$ | |||||||||||
\(\ds z_2\) | \(=\) | \(\ds \tuple {x_2, y_2}\) | Definition 2 of Complex Number: for some $x_2, y_2 \in \R$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z_1 z_2\) | \(=\) | \(\ds \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2}\) | Definition of Complex Multiplication | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds x_1 x_2\) | \(=\) | \(\ds y_1 y_2\) | ||||||||||
\(\text {(2)}: \quad\) | \(\, \ds \land \, \) | \(\ds x_1 y_2\) | \(=\) | \(\ds -y_1 x_2\) | as $z_1 z_2 = 0$ |
Without loss of generality, let $\tuple {x_2, y_2} \ne \tuple {0, 0}$.
Aiming for a contradiction, suppose also that $\tuple {x_1, y_1} \ne \tuple {0, 0}$.
Then:
\(\ds x_1\) | \(=\) | \(\ds \frac {y_1 y_2} {x_2}\) | from $(1)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {y_1 y_2} {x_2} y_2\) | \(=\) | \(\ds -y_1 x_2\) | substituting in $(2)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y_2^2\) | \(=\) | \(\ds - x_2^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_2 = y_2\) | \(=\) | \(\ds 0\) | as both $x_2$ and $y_2$ are real |
From this contradiction it follows that:
- $\tuple {x_1, y_1} = \tuple {0, 0}$
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $75$