Question
Answer
$$\dfrac{4-2i}{2i(i-1)^2}=\dfrac{4-2i}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4-2i}{2i(i-1)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4-2i}{2i(1i^2-2i+1)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4-2i}{2i(-1-2i+1)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-2i}{2i\cdot-2i} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{4-2i}{-4i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{4-2i}{4}\end{aligned} $$ | |
| ① | Find $ \left(i-1\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ i } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(i-1\right)^2 = \color{blue}{i^2} -2 \cdot i \cdot 1 + \color{red}{1^2} = i^2-2i+1\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ -\cancel{1}} \,-2i+ \, \color{blue}{ \cancel{1}} \, = -2i $$ |
| ④ | $$ -4i^2 = -4 \cdot (-1) = 4 $$ |
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