Question
Answer
$$(7-7i)\cdot(4-3i)=21i^2-49i+28$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(7-7i)\cdot(4-3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}28-21i-28i+21i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}21i^2-49i+28\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{7-7i}\right) $ by each term in $ \left( 4-3i\right) $. $$ \left( \color{blue}{7-7i}\right) \cdot \left( 4-3i\right) = 28-21i-28i+21i^2 $$ |
| ② | Combine like terms: $$ 28 \color{blue}{-21i} \color{blue}{-28i} +21i^2 = 21i^2 \color{blue}{-49i} +28 $$ |
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