Question
Answer
$$(1+2j)\cdot(1-3j)=-6j^2-j+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+2j)\cdot(1-3j)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1-3j+2j-6j^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-6j^2-j+1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+2j}\right) $ by each term in $ \left( 1-3j\right) $. $$ \left( \color{blue}{1+2j}\right) \cdot \left( 1-3j\right) = 1-3j+2j-6j^2 $$ |
| ② | Combine like terms: $$ 1 \color{blue}{-3j} + \color{blue}{2j} -6j^2 = -6j^2 \color{blue}{-j} +1 $$ |
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