Answer
$$\frac{i}{i+1}-(1-i)^2=\frac{5i+1}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{i}{i+1}-(1-i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{i}{i+1}-(1-2i+i^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{i}{i+1}-(1-2i-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{i}{i+1}--2i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1+i}{2}--2i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{5i+1}{2}\end{aligned} $$ | |
| ① | 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 }$. $$ \begin{aligned}\left(1-i\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot i + \color{red}{i^2} = 1-2i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,-2i \, \color{blue}{ -\cancel{1}} \, = -2i $$ |
| ④ | Divide $ \, i \, $ by $ \, 1+i \, $ to get $\,\, \dfrac{1+i}{2} $. ( view steps ) |
| ⑤ | Subtract $-2i$ from $ \dfrac{1+i}{2} $ to get $ \dfrac{ \color{purple}{ 5i+1 } }{ 2 }$. Step 1: Write $ -2i $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |
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