Tap the blue circles to see an explanation.
| $$ \begin{aligned}-12i\frac{i}{(i-1)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12i\frac{i}{i^2-2i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12i\frac{i}{-1-2i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}12i\frac{i}{-2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}12i(-\frac{1}{2}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{12i}{2}\end{aligned} $$ | |
| 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 | Divide $ \, i \, $ by $ \, -2i \, $ to get $\,\, \dfrac{-1}{2} $. ( view steps ) |
| ⑤ | Multiply $12i$ by $ \dfrac{-1}{2} $ to get $ \dfrac{ -12i }{ 2 } $. Step 1: Write $ 12i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 12i \cdot \frac{-1}{2} & \xlongequal{\text{Step 1}} \frac{12i}{\color{red}{1}} \cdot \frac{-1}{2} \xlongequal{\text{Step 2}} \frac{ 12i \cdot \left( -1 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -12i }{ 2 } \end{aligned} $$ |