Question
Answer
$$(3-4i)\cdot(1+5i)=-20i^2+11i+3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-4i)\cdot(1+5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3+15i-4i-20i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-20i^2+11i+3\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-4i}\right) $ by each term in $ \left( 1+5i\right) $. $$ \left( \color{blue}{3-4i}\right) \cdot \left( 1+5i\right) = 3+15i-4i-20i^2 $$ |
| ② | Combine like terms: $$ 3+ \color{blue}{15i} \color{blue}{-4i} -20i^2 = -20i^2+ \color{blue}{11i} +3 $$ |
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