# How do you multiply imaginary numbers of the form a + bi?

Nov 29, 2015

See explanation

#### Explanation:

$\textcolor{b l u e}{\text{Comment}}$

Example
$\left(\text{Real part" + "Imaginary part}\right) \to \left(R e + I m\right) \to \left(6 + 12 i\right)$
Notice that I keep the brackets. This is important as the 'Real' part,6, contributes to the whole. As does the Imaginary part of 12i. So together they are the whole of something.

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The distributive law applies

$\textcolor{b l u e}{\text{Case 1}} \textcolor{w h i t e}{\ldots .} c \times \left(a + b i\right)$

Write as : $\left(c \times a\right) + \left(c \times b\right) i \to \left(c a + c b i\right)$

Using the actual numbers previously chosen at random:

$3 \left(2 + 6 i\right) \to \left(6 + 12 i\right)$

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$\textcolor{b l u e}{\text{Case 2}} \textcolor{w h i t e}{\ldots .} \left(a + b i\right) \left(c + \mathrm{di}\right)$

This would be easier to demonstrate with numbers

Suppose we had: $\left(2 + i\right) \left(3 + 4 i\right)$

$2 \left(3 + 4 i\right) + i \left(3 + 4 i\right)$

$6 + 8 i + 3 i + 4 {i}^{2}$

$6 + 11 i + 4 {i}^{2}$

But $i = \sqrt{- 1} \text{ so } {i}^{2} = \left(- 1\right)$

$6 + 11 i + 4 \left(- 1\right)$

$2 + 11 i$
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