Question
Answer
$$(2-3i)\cdot(4-5i)=15i^2-22i+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2-3i)\cdot(4-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}8-10i-12i+15i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}15i^2-22i+8\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{2-3i}\right) $ by each term in $ \left( 4-5i\right) $. $$ \left( \color{blue}{2-3i}\right) \cdot \left( 4-5i\right) = 8-10i-12i+15i^2 $$ |
| ② | Combine like terms: $$ 8 \color{blue}{-10i} \color{blue}{-12i} +15i^2 = 15i^2 \color{blue}{-22i} +8 $$ |
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