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