Question
Answer
$$(1+4i)\cdot(7+3i)=12i^2+31i+7$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+4i)\cdot(7+3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}7+3i+28i+12i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12i^2+31i+7\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+4i}\right) $ by each term in $ \left( 7+3i\right) $. $$ \left( \color{blue}{1+4i}\right) \cdot \left( 7+3i\right) = 7+3i+28i+12i^2 $$ |
| ② | Combine like terms: $$ 7+ \color{blue}{3i} + \color{blue}{28i} +12i^2 = 12i^2+ \color{blue}{31i} +7 $$ |
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