Real Number Multiplied by Complex Number

From ProofWiki
Jump to navigation Jump to search


Let $a \in \R$ be a real number.

Let $c + d i \in \C$ be a complex number.


$a \times \left({c + d i}\right) = \left({c + d i}\right) \times a = a c + i a d$


$a$ can be expressed as a wholly real complex number $a + 0 i$.

Then we have:

\(\ds a \times \left({c + d i}\right)\) \(=\) \(\ds \left({a + 0 i}\right) \times \left({c + d i}\right)\)
\(\ds \) \(=\) \(\ds \left({a c - 0 d}\right) + \left({a d + 0 c}\right) i\)
\(\ds \) \(=\) \(\ds a c + i a d\)

The result for $\left({c + d i}\right) \times a$ follows from Complex Multiplication is Commutative.