Question
Answer
$$(4+3i)\cdot(4-3i)=-9i^2+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4+3i)\cdot(4-3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}16-12i+12i-9i^2 \xlongequal{ } \\[1 em] & \xlongequal{ }16 -\cancel{12i}+ \cancel{12i}-9i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-9i^2+16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4+3i}\right) $ by each term in $ \left( 4-3i\right) $. $$ \left( \color{blue}{4+3i}\right) \cdot \left( 4-3i\right) = 16 -\cancel{12i}+ \cancel{12i}-9i^2 $$ |
| ② | Combine like terms: $$ 16 \, \color{blue}{ -\cancel{12i}} \,+ \, \color{blue}{ \cancel{12i}} \,-9i^2 = -9i^2+16 $$ |
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