Question
Answer
$$(1+2i)\cdot(3+2i)+(5-2i)^2=-12i+20$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+2i)\cdot(3+2i)+(5-2i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1+2i)\cdot(3+2i)+25-20i+4i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1+2i)\cdot(3+2i)+25-20i-4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1+2i)\cdot(3+2i)-20i+21 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}3+2i+6i+4i^2-20i+21 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4i^2+8i+3-20i+21 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-4+8i+3-20i+21 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}8i-1-20i+21 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}-12i+20\end{aligned} $$ | |
| ① | Find $ \left(5-2i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 5 } $ and $ B = \color{red}{ 2i }$. $$ \begin{aligned}\left(5-2i\right)^2 = \color{blue}{5^2} -2 \cdot 5 \cdot 2i + \color{red}{\left( 2i \right)^2} = 25-20i+4i^2\end{aligned} $$ |
| ② | $$ 4i^2 = 4 \cdot (-1) = -4 $$ |
| ③ | Combine like terms: $$ \color{blue}{25} -20i \color{blue}{-4} = -20i+ \color{blue}{21} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{1+2i}\right) $ by each term in $ \left( 3+2i\right) $. $$ \left( \color{blue}{1+2i}\right) \cdot \left( 3+2i\right) = 3+2i+6i+4i^2 $$ |
| ⑤ | Combine like terms: $$ 3+ \color{blue}{2i} + \color{blue}{6i} +4i^2 = 4i^2+ \color{blue}{8i} +3 $$ |
| ⑥ | $$ 4i^2 = 4 \cdot (-1) = -4 $$ |
| ⑦ | Combine like terms: $$ \color{blue}{-4} +8i+ \color{blue}{3} = 8i \color{blue}{-1} $$ |
| ⑧ | Combine like terms: $$ \color{blue}{8i} \color{red}{-1} \color{blue}{-20i} + \color{red}{21} = \color{blue}{-12i} + \color{red}{20} $$ |
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