Question
Answer
$$(11-9i)\cdot(11+9i)=-81i^2+121$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(11-9i)\cdot(11+9i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}121+99i-99i-81i^2 \xlongequal{ } \\[1 em] & \xlongequal{ }121+ \cancel{99i} -\cancel{99i}-81i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-81i^2+121\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{11-9i}\right) $ by each term in $ \left( 11+9i\right) $. $$ \left( \color{blue}{11-9i}\right) \cdot \left( 11+9i\right) = 121+ \cancel{99i} -\cancel{99i}-81i^2 $$ |
| ② | Combine like terms: $$ 121+ \, \color{blue}{ \cancel{99i}} \, \, \color{blue}{ -\cancel{99i}} \,-81i^2 = -81i^2+121 $$ |
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