Answer
$$\frac{(1+i)^2+(1-i)^2}{(1+i)^2-(1-i)^2}=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(1+i)^2+(1-i)^2}{(1+i)^2-(1-i)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1+2i+i^2+1-2i+i^2}{1+2i+i^2-(1-2i+i^2)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{1+2i-1+1-2i-1}{1+2i-1-(1-2i-1)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{2i-2i}{2i--2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle13}{\textcircled {13}} } }}}\frac{0}{2i+2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle14}{\textcircled {14}} } }}}\frac{0}{4i} \xlongequal{ } \\[1 em] & \xlongequal{ }0\end{aligned} $$ | |
| 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 | 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 }$. $$ \begin{aligned}\left(1-i\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot i + \color{red}{i^2} = 1-2i+i^2\end{aligned} $$ |
| ③ | 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 }$. $$ \begin{aligned}\left(1+i\right)^2 = \color{blue}{1^2} +2 \cdot 1 \cdot i + \color{red}{i^2} = 1+2i+i^2\end{aligned} $$ |
| ④ | 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 }$. $$ \begin{aligned}\left(1-i\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot i + \color{red}{i^2} = 1-2i+i^2\end{aligned} $$ |
| ⑤ | $$ i^2 = -1 $$ |
| ⑥ | $$ i^2 = -1 $$ |
| ⑦ | $$ i^2 = -1 $$ |
| ⑧ | $$ i^2 = -1 $$ |
| ⑨ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,+2i \, \color{blue}{ -\cancel{1}} \, = 2i $$ |
| ⑩ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,-2i \, \color{blue}{ -\cancel{1}} \, = -2i $$ |
| ⑪ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,+2i \, \color{blue}{ -\cancel{1}} \, = 2i $$ |
| ⑫ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,-2i \, \color{blue}{ -\cancel{1}} \, = -2i $$ |
| ⑬ | Simplify numerator $$ \, \color{blue}{ \cancel{2i}} \, \, \color{blue}{ -\cancel{2i}} \, = 0 $$ |
| ⑭ | Simplify denominator $$ \color{blue}{2i} + \color{blue}{2i} = \color{blue}{4i} $$ |
This page was created using
Complex Expression calculator