Question
Answer
$$(3-7i)\cdot(3+7i)=-49i^2+9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-7i)\cdot(3+7i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9+21i-21i-49i^2 \xlongequal{ } \\[1 em] & \xlongequal{ }9+ \cancel{21i} -\cancel{21i}-49i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-49i^2+9\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-7i}\right) $ by each term in $ \left( 3+7i\right) $. $$ \left( \color{blue}{3-7i}\right) \cdot \left( 3+7i\right) = 9+ \cancel{21i} -\cancel{21i}-49i^2 $$ |
| ② | Combine like terms: $$ 9+ \, \color{blue}{ \cancel{21i}} \, \, \color{blue}{ -\cancel{21i}} \,-49i^2 = -49i^2+9 $$ |
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