Answer
$$\frac{(i+1)^2}{(i-1)^4}=\frac{2i}{-4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(i+1)^2}{(i-1)^4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{i^2+2i+1}{i^4-4i^3+6i^2-4i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-1+2i+1}{i^4-4i^3+6i^2-4i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-1+2i+1}{1+4i-6-4i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{2i}{1+4i-6-4i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{2i}{-4}\end{aligned} $$ | |
| 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 | $$ (i-1)^4 = (i-1)^2 \cdot (i-1)^2 $$ |
| ③ | 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 }$. $$ \begin{aligned}\left(i-1\right)^2 = \color{blue}{i^2} -2 \cdot i \cdot 1 + \color{red}{1^2} = i^2-2i+1\end{aligned} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{i^2-2i+1}\right) $ by each term in $ \left( i^2-2i+1\right) $. $$ \left( \color{blue}{i^2-2i+1}\right) \cdot \left( i^2-2i+1\right) = i^4-2i^3+i^2-2i^3+4i^2-2i+i^2-2i+1 $$ |
| ⑤ | Combine like terms: $$ i^4 \color{blue}{-2i^3} + \color{red}{i^2} \color{blue}{-2i^3} + \color{green}{4i^2} \color{orange}{-2i} + \color{green}{i^2} \color{orange}{-2i} +1 = i^4 \color{blue}{-4i^3} + \color{green}{6i^2} \color{orange}{-4i} +1 $$ |
| ⑥ | $$ i^2 = -1 $$ |
| ⑦ | $$ i^4 = i^2 \cdot i^2 =
( - 1) \cdot ( - 1) =
1 $$$$ -4i^3 = -4 \cdot \color{blue}{i^2} \cdot i =
-4 \cdot ( \color{blue}{-1}) \cdot i =
4 \cdot \, i $$$$ 6i^2 = 6 \cdot (-1) = -6 $$ |
| ⑧ | Simplify numerator $$ \, \color{blue}{ -\cancel{1}} \,+2i+ \, \color{blue}{ \cancel{1}} \, = 2i $$ |
| ⑨ | Simplify denominator $$ \color{blue}{1} + \, \color{red}{ \cancel{4i}} \, \color{orange}{-6} \, \color{red}{ -\cancel{4i}} \,+ \color{orange}{1} = \color{orange}{-4} $$ |
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