Question
Answer
$$(1+7i)\cdot(9+3i)-(4+2i)=64i-16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+7i)\cdot(9+3i)-(4+2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9+3i+63i+21i^2-(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}21i^2+66i+9-(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-21+66i+9-(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}66i-12-(4+2i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}66i-12-4-2i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}64i-16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+7i}\right) $ by each term in $ \left( 9+3i\right) $. $$ \left( \color{blue}{1+7i}\right) \cdot \left( 9+3i\right) = 9+3i+63i+21i^2 $$ |
| ② | Combine like terms: $$ 9+ \color{blue}{3i} + \color{blue}{63i} +21i^2 = 21i^2+ \color{blue}{66i} +9 $$ |
| ③ | $$ 21i^2 = 21 \cdot (-1) = -21 $$ |
| ④ | Combine like terms: $$ \color{blue}{-21} +66i+ \color{blue}{9} = 66i \color{blue}{-12} $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4+2i \right) = -4-2i $$ |
| ⑥ | Combine like terms: $$ \color{blue}{66i} \color{red}{-12} \color{red}{-4} \color{blue}{-2i} = \color{blue}{64i} \color{red}{-16} $$ |
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