Complex Addition is Closed

Theorem

The set of complex numbers $\C$ is closed under addition:

$\forall z, w \in \C: z + w \in \C$

Proof

Proof from Informal Definition

From the informal definition of complex numbers, we define the following:

• $z = x_1 + i y_1$
• $w = x_2 + i y_2$

where $i = \sqrt {-1}$ and $x_1, x_2, y_1, y_2 \in \R$.

Then from the definition of complex addition, $z + w = \left({x_1 + x_2}\right) + i \left({y_1 + y_2}\right)$.

Real Addition is closed, from Additive Group of Real Numbers so $\left({x_1 + x_2}\right) \in \R$ and $\left({y_1 + y_2}\right) \in \R$.

Hence the result.

$\blacksquare$

Proof from Formal Definition

From the formal definition of complex numbers, we have:

• $z = \left({x_1, y_1}\right)$
• $w = \left({x_2, y_2}\right)$

where $x_1, x_2, y_1, y_2 \in \R$.

Then from the definition of complex addition, $z + w = \left({x_1 + x_2, y_1 + y_2}\right)$.

Real Addition is closed, from Additive Group of Real Numbers so $\left({x_1 + x_2}\right) \in \R$ and $\left({y_1 + y_2}\right) \in \R$.

Hence the result.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 1.2$: Example $1$