Real Number Multiplied by Complex Number

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Theorem

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

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


Then:

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


Proof

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

Then we have:

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


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

$\blacksquare$


Sources