Answer
$$(8+4i)\cdot(11-3i)=-12i^2+20i+88$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(8+4i)\cdot(11-3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}88-24i+44i-12i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-12i^2+20i+88\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{8+4i}\right) $ by each term in $ \left( 11-3i\right) $. $$ \left( \color{blue}{8+4i}\right) \cdot \left( 11-3i\right) = 88-24i+44i-12i^2 $$ |
| ② | Combine like terms: $$ 88 \color{blue}{-24i} + \color{blue}{44i} -12i^2 = -12i^2+ \color{blue}{20i} +88 $$ |
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