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