Question
Answer
$$2(2-3i)^2+2-3i-3=-27i-11$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2(2-3i)^2+2-3i-3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2(4-12i+9i^2)+2-3i-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2(4-12i-9)+2-3i-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2(-12i-5)+2-3i-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-24i-10+2-3i-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-27i-8-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-27i-11\end{aligned} $$ | |
| ① | Find $ \left(2-3i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2 } $ and $ B = \color{red}{ 3i }$. $$ \begin{aligned}\left(2-3i\right)^2 = \color{blue}{2^2} -2 \cdot 2 \cdot 3i + \color{red}{\left( 3i \right)^2} = 4-12i+9i^2\end{aligned} $$ |
| ② | $$ 9i^2 = 9 \cdot (-1) = -9 $$ |
| ③ | Combine like terms: $$ \color{blue}{4} -12i \color{blue}{-9} = -12i \color{blue}{-5} $$ |
| ④ | Multiply $ \color{blue}{2} $ by $ \left( -12i-5\right) $ $$ \color{blue}{2} \cdot \left( -12i-5\right) = -24i-10 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-24i} \color{red}{-10} + \color{red}{2} \color{blue}{-3i} = \color{blue}{-27i} \color{red}{-8} $$ |
| ⑥ | Combine like terms: $$ -27i \color{blue}{-8} \color{blue}{-3} = -27i \color{blue}{-11} $$ |
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