Question
Answer
$$((-1+i)^2+2\cdot(-1+i)+3)^2=1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}((-1+i)^2+2\cdot(-1+i)+3)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1-2i+i^2+2\cdot(-1+i)+3)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1-2i-1+2\cdot(-1+i)+3)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(-2i+2\cdot(-1+i)+3)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(-2i-2+2i+3)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}(-2+3)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}1^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}1\end{aligned} $$ | |
| ① | Find $ \left(-1+i\right)^2 $ in two steps. S1: Change all signs inside bracket. S2: Apply 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 }$. $$ \begin{aligned}\left(-1+i\right)^2& \xlongequal{ S1 } \left(1-i\right)^2 \xlongequal{ S2 } \color{blue}{1^2} -2 \cdot 1 \cdot i + \color{red}{i^2} = \\[1 em] & = 1-2i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,-2i \, \color{blue}{ -\cancel{1}} \, = -2i $$ |
| ④ | Multiply $ \color{blue}{2} $ by $ \left( -1+i\right) $ $$ \color{blue}{2} \cdot \left( -1+i\right) = -2+2i $$ |
| ⑤ | Combine like terms: $$ \, \color{blue}{ -\cancel{2i}} \,-2+ \, \color{blue}{ \cancel{2i}} \, = -2 $$ |
| ⑥ | Combine like terms: $$ \color{blue}{-2} + \color{blue}{3} = \color{blue}{1} $$ |
| ⑦ | -2+3=1 |
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