Question
Answer
$$(1+2i+3-i)\cdot(2+i)=i^2+6i+8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+2i+3-i)\cdot(2+i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1i+4)\cdot(2+i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2i+i^2+8+4i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}i^2+6i+8\end{aligned} $$ | |
| ① | Combine like terms: $$ \color{blue}{1} + \color{red}{2i} + \color{blue}{3} \color{red}{-i} = \color{red}{i} + \color{blue}{4} $$ |
| ② | Multiply each term of $ \left( \color{blue}{i+4}\right) $ by each term in $ \left( 2+i\right) $. $$ \left( \color{blue}{i+4}\right) \cdot \left( 2+i\right) = 2i+i^2+8+4i $$ |
| ③ | Combine like terms: $$ \color{blue}{2i} +i^2+8+ \color{blue}{4i} = i^2+ \color{blue}{6i} +8 $$ |
This page was created using
Complex Expression calculator