Question
Answer
$$\dfrac{2i^3+1}{(2+i)(1-i)^2}=\dfrac{1}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2i^3+1}{(2+i)(1-i)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-2i+1}{(2+i)(1-2i+i^2)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2i+1}{(2+i)(1-2i-1)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-2i+1}{(2+i)\cdot-2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-2i+1}{-4i-2i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{-2i+1}{-4i+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{1}{2}\end{aligned} $$ | |
| ① | $$ 2i^3 = 2 \cdot \color{blue}{i^2} \cdot i =
2 \cdot ( \color{blue}{-1}) \cdot i =
-2 \cdot \, i $$ |
| ② | Find $ \left(1-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}{ 1 } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(1-i\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot i + \color{red}{i^2} = 1-2i+i^2\end{aligned} $$ |
| ③ | $$ 2i^3 = 2 \cdot \color{blue}{i^2} \cdot i =
2 \cdot ( \color{blue}{-1}) \cdot i =
-2 \cdot \, i $$ |
| ④ | $$ i^2 = -1 $$ |
| ⑤ | $$ 2i^3 = 2 \cdot \color{blue}{i^2} \cdot i =
2 \cdot ( \color{blue}{-1}) \cdot i =
-2 \cdot \, i $$ |
| ⑥ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,-2i \, \color{blue}{ -\cancel{1}} \, = -2i $$ |
| ⑦ | $$ \left( \color{blue}{2+i}\right) \cdot -2i = -4i-2i^2 $$ |
| ⑧ | $$ -2i^2 = -2 \cdot (-1) = 2 $$ |
| ⑨ | Divide $ \, 1-2i \, $ by $ \, 2-4i \, $ to get $\,\, \dfrac{1}{2} $. ( view steps ) |
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