Answer
$$(11-i)\cdot(2-i)=i^2-13i+22$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(11-i)\cdot(2-i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}22-11i-2i+i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}i^2-13i+22\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{11-i}\right) $ by each term in $ \left( 2-i\right) $. $$ \left( \color{blue}{11-i}\right) \cdot \left( 2-i\right) = 22-11i-2i+i^2 $$ |
| ② | Combine like terms: $$ 22 \color{blue}{-11i} \color{blue}{-2i} +i^2 = i^2 \color{blue}{-13i} +22 $$ |
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