Question
Answer
$$(1+2i)\cdot(2+3i)\cdot(3+4i)=28i^2+5i-12$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+2i)\cdot(2+3i)\cdot(3+4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(2+3i+4i+6i^2)\cdot(3+4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(6i^2+7i+2)\cdot(3+4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(-6+7i+2)\cdot(3+4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(7i-4)\cdot(3+4i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}21i+28i^2-12-16i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}28i^2+5i-12\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+2i}\right) $ by each term in $ \left( 2+3i\right) $. $$ \left( \color{blue}{1+2i}\right) \cdot \left( 2+3i\right) = 2+3i+4i+6i^2 $$ |
| ② | Combine like terms: $$ 2+ \color{blue}{3i} + \color{blue}{4i} +6i^2 = 6i^2+ \color{blue}{7i} +2 $$ |
| ③ | $$ 6i^2 = 6 \cdot (-1) = -6 $$ |
| ④ | Combine like terms: $$ \color{blue}{-6} +7i+ \color{blue}{2} = 7i \color{blue}{-4} $$ |
| ⑤ | Multiply each term of $ \left( \color{blue}{7i-4}\right) $ by each term in $ \left( 3+4i\right) $. $$ \left( \color{blue}{7i-4}\right) \cdot \left( 3+4i\right) = 21i+28i^2-12-16i $$ |
| ⑥ | Combine like terms: $$ \color{blue}{21i} +28i^2-12 \color{blue}{-16i} = 28i^2+ \color{blue}{5i} -12 $$ |
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