Question
Answer
$$(1+i)^2-2\cdot(1+i)+2=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+i)^2-2\cdot(1+i)+2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1+2i+i^2-2\cdot(1+i)+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}1+2i-1-2\cdot(1+i)+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2i-2\cdot(1+i)+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2i-(2+2i)+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2i-2-2i+2 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{2i} -\cancel{2} -\cancel{2i}+ \cancel{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}0\end{aligned} $$ | |
| 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 | Multiply $ \color{blue}{2} $ by $ \left( 1+i\right) $ $$ \color{blue}{2} \cdot \left( 1+i\right) = 2+2i $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 2+2i \right) = -2-2i $$ |
| ⑥ | Combine like terms: $$ \, \color{blue}{ \cancel{2i}} \, \, \color{green}{ -\cancel{2}} \, \, \color{blue}{ -\cancel{2i}} \,+ \, \color{green}{ \cancel{2}} \, = \color{green}{0} $$ |
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