Question
Answer
$$(4+5i)\cdot(-2+3i)=15i^2+2i-8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4+5i)\cdot(-2+3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-8+12i-10i+15i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}15i^2+2i-8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4+5i}\right) $ by each term in $ \left( -2+3i\right) $. $$ \left( \color{blue}{4+5i}\right) \cdot \left( -2+3i\right) = -8+12i-10i+15i^2 $$ |
| ② | Combine like terms: $$ -8+ \color{blue}{12i} \color{blue}{-10i} +15i^2 = 15i^2+ \color{blue}{2i} -8 $$ |
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