Question
Answer
$$\dfrac{8+2i-(1-i)}{(2+i)^2}=\dfrac{33-19i}{25}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{8+2i-(1-i)}{(2+i)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8+2i-1+i}{4+4i+i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8+2i-1+i}{4+4i-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{3i+7}{4+4i-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{3i+7}{4i+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{33-19i}{25}\end{aligned} $$ | |
| ① | Remove the parentheses by changing the sign of each term within them. $$ - \left( 1-i \right) = -1+i $$ |
| ② | 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 }$. $$ \begin{aligned}\left(2+i\right)^2 = \color{blue}{2^2} +2 \cdot 2 \cdot i + \color{red}{i^2} = 4+4i+i^2\end{aligned} $$ |
| ③ | $$ i^2 = -1 $$ |
| ④ | Simplify numerator $$ \color{blue}{8} + \color{red}{2i} \color{blue}{-1} + \color{red}{i} = \color{red}{3i} + \color{blue}{7} $$ |
| ⑤ | Simplify denominator $$ \color{blue}{4} +4i \color{blue}{-1} = 4i+ \color{blue}{3} $$ |
| ⑥ | Divide $ \, 7+3i \, $ by $ \, 3+4i \, $ to get $\,\, \dfrac{33-19i}{25} $. ( view steps ) |
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