Question
Answer
$$(-11+3i)\cdot(9+2i)=6i^2+5i-99$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-11+3i)\cdot(9+2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-99-22i+27i+6i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}6i^2+5i-99\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{-11+3i}\right) $ by each term in $ \left( 9+2i\right) $. $$ \left( \color{blue}{-11+3i}\right) \cdot \left( 9+2i\right) = -99-22i+27i+6i^2 $$ |
| ② | Combine like terms: $$ -99 \color{blue}{-22i} + \color{blue}{27i} +6i^2 = 6i^2+ \color{blue}{5i} -99 $$ |
This page was created using
Complex Expression calculator