Complex Addition is Closed

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Theorem

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

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


Proof 1

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:

$z + w = \left({x_1 + x_2}\right) + i \left({y_1 + y_2}\right)$

From Real Numbers under Addition form Abelian Group, real addition is closed.

So:

$\left({x_1 + x_2}\right) \in \R$ and $\left({y_1 + y_2}\right) \in \R$

Hence the result.

$\blacksquare$


Proof 2

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:

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

From Real Numbers under Addition form Abelian Group, real addition is closed.

So: $\left({x_1 + x_2}\right) \in \R$ and $\left({y_1 + y_2}\right) \in \R$

Hence the result.

$\blacksquare$


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