Question
Answer
$$\dfrac{(2+i)^2\cdot(3-i)}{2-3i}=\dfrac{-1+57i}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(2+i)^2\cdot(3-i)}{2-3i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{(4+4i+i^2)\cdot(3-i)}{2-3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{(4+4i-1)\cdot(3-i)}{2-3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{(4i+3)\cdot(3-i)}{2-3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{12i-4i^2+9-3i}{2-3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{12i+4+9-3i}{2-3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{9i+13}{2-3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-1+57i}{13}\end{aligned} $$ | |
| ① | Find $ \left(2+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}{ 2 } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(2+i\right)^2 = \color{blue}{2^2} +2 \cdot 2 \cdot i + \color{red}{i^2} = 4+4i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \color{blue}{4} +4i \color{blue}{-1} = 4i+ \color{blue}{3} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{4i+3}\right) $ by each term in $ \left( 3-i\right) $. $$ \left( \color{blue}{4i+3}\right) \cdot \left( 3-i\right) = 12i-4i^2+9-3i $$ |
| ⑤ | $$ -4i^2 = -4 \cdot (-1) = 4 $$ |
| ⑥ | Simplify numerator $$ \color{blue}{12i} + \color{red}{4} + \color{red}{9} \color{blue}{-3i} = \color{blue}{9i} + \color{red}{13} $$ |
| ⑦ | Divide $ \, 13+9i \, $ by $ \, 2-3i \, $ to get $\,\, \dfrac{-1+57i}{13} $. ( view steps ) |
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