Question
Answer
$$\dfrac{3(a+bi)^2-3}{(a+bi)^2+4(a+bi)+1}=\dfrac{3a^2+6abi+3b^2i^2-3}{b^2i^2+2abi+a^2+4bi+4a+1}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3(a+bi)^2-3}{(a+bi)^2+4(a+bi)+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{((a+bi)^2-1)\cdot3}{a^2+2abi+b^2i^2+4a+4bi+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{(1a^2+2abi+b^2i^2-1)\cdot3}{b^2i^2+2abi+a^2+4bi+4a+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{3a^2+6abi+3b^2i^2-3}{b^2i^2+2abi+a^2+4bi+4a+1}\end{aligned} $$ | |
| ① | Use the distributive property. |
| ② | Find $ \left(a+bi\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ a } $ and $ B = \color{red}{ bi }$. $$ \begin{aligned}\left(a+bi\right)^2 = \color{blue}{a^2} +2 \cdot a \cdot bi + \color{red}{\left( bi \right)^2} = a^2+2abi+b^2i^2\end{aligned} $$ |
| ③ | Multiply $ \color{blue}{4} $ by $ \left( a+bi\right) $ $$ \color{blue}{4} \cdot \left( a+bi\right) = 4a+4bi $$ |
| ④ | Find $ \left(a+bi\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ a } $ and $ B = \color{red}{ bi }$. $$ \begin{aligned}\left(a+bi\right)^2 = \color{blue}{a^2} +2 \cdot a \cdot bi + \color{red}{\left( bi \right)^2} = a^2+2abi+b^2i^2\end{aligned} $$ |
| ⑤ | Combine like terms: $$ a^2+2abi+b^2i^2+4a+4bi = b^2i^2+2abi+a^2+4bi+4a $$ |
| ⑥ | $$ \left( \color{blue}{a^2+2abi+b^2i^2-1}\right) \cdot 3 = 3a^2+6abi+3b^2i^2-3 $$ |
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