Question
Answer
$$\dfrac{i(x+iy)^2-2(x+iy)}{3(x+iy)+1-i}=\dfrac{ix^2+2i^2xy+i^3y^2-2x-2iy}{3x+3iy+1-i}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{i(x+iy)^2-2(x+iy)}{3(x+iy)+1-i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{i(x^2+2ixy+i^2y^2)-2(x+iy)}{3x+3iy+1-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{ix^2+2i^2xy+i^3y^2-(2x+2iy)}{3x+3iy+1-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{ix^2+2i^2xy+i^3y^2-2x-2iy}{3x+3iy+1-i}\end{aligned} $$ | |
| ① | Find $ \left(x+iy\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ iy }$. $$ \begin{aligned}\left(x+iy\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot iy + \color{red}{\left( iy \right)^2} = x^2+2ixy+i^2y^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{3} $ by $ \left( x+iy\right) $ $$ \color{blue}{3} \cdot \left( x+iy\right) = 3x+3iy $$ |
| ③ | Multiply $ \color{blue}{i} $ by $ \left( x^2+2ixy+i^2y^2\right) $ $$ \color{blue}{i} \cdot \left( x^2+2ixy+i^2y^2\right) = ix^2+2i^2xy+i^3y^2 $$ |
| ④ | Multiply $ \color{blue}{2} $ by $ \left( x+iy\right) $ $$ \color{blue}{2} \cdot \left( x+iy\right) = 2x+2iy $$ |
| ⑤ | Multiply $ \color{blue}{3} $ by $ \left( x+iy\right) $ $$ \color{blue}{3} \cdot \left( x+iy\right) = 3x+3iy $$ |
| ⑥ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 2x+2iy \right) = -2x-2iy $$ |
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