# Non-Zero Complex Numbers are Closed under Multiplication

## Theorem

The set of non-zero complex numbers is closed under multiplication.

## Proof 1

Recall that Complex Numbers form Field under the operations of addition and multiplication.

By definition of a field, the algebraic structure $\struct {\C_{\ne 0}, \times}$ is a group.

Thus, by definition, $\times$ is closed in $\struct {\C_{\ne 0}, \times}$.

$\blacksquare$

## Proof 2

Let $z_1, z_2 \in \C_{\ne 0}$.

Then by definition of complex number:

- $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$

for some $x_1, y_1, x_2, y_2 \in \R$ such that:

- $x_1 \ne 0$ or $y_1 \ne 0$
- $x_2 \ne 0$ or $y_2 \ne 0$

Expressing $z_1$ and $z_2$ in exponential form (although polar form is equally adequate):

- $z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$

for some $r_1, r_2, \theta_1, \theta_2 \in \R$.

Then by Product of Complex Numbers in Polar Form:

- $z_1 \times z_2 = \paren {r_1 \times r_2} e^{i \paren {\theta_1 + \theta_2} }$

By definition of exponential form:

- $r_1 = \sqrt {x_1^2 + y_1^2}$
- $r_2 = \sqrt {x_2^2 + y_2^2}$

Thus $r_1 > 0$ and $r_2 > 0$.

Hence $r_1 \times r_2 > 0$ and so $z_1 \times z_2 \ne 0$.

$\blacksquare$

## Proof 3

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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.2$