Question
Answer
$$(3-i)^2\cdot(5-2i)+8i=12i^2-38i+40$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-i)^2\cdot(5-2i)+8i& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(9-6i+i^2)\cdot(5-2i)+8i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(9-6i-1)\cdot(5-2i)+8i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(-6i+8)\cdot(5-2i)+8i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-30i+12i^2+40-16i+8i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}12i^2-38i+40\end{aligned} $$ | |
| ① | Find $ \left(3-i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3 } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(3-i\right)^2 = \color{blue}{3^2} -2 \cdot 3 \cdot i + \color{red}{i^2} = 9-6i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \color{blue}{9} -6i \color{blue}{-1} = -6i+ \color{blue}{8} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{-6i+8}\right) $ by each term in $ \left( 5-2i\right) $. $$ \left( \color{blue}{-6i+8}\right) \cdot \left( 5-2i\right) = -30i+12i^2+40-16i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-30i} +12i^2+40 \color{red}{-16i} + \color{red}{8i} = 12i^2 \color{red}{-38i} +40 $$ |
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