Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{c(a+ib)-(a+ib)^2+d}{c(a+ib)+(a+ib)^2-d}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{ac+bci-(1a^2+2abi+b^2i^2)+d}{ac+bci+a^2+2abi+b^2i^2-d} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{ac+bci-a^2-2abi-b^2i^2+d}{b^2i^2+2abi+bci+a^2+ac-d}\end{aligned} $$ | |
① | Multiply $ \color{blue}{c} $ by $ \left( a+bi\right) $ $$ \color{blue}{c} \cdot \left( a+bi\right) = ac+bci $$ |
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}{c} $ by $ \left( a+bi\right) $ $$ \color{blue}{c} \cdot \left( a+bi\right) = ac+bci $$ |
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 | Remove the parentheses by changing the sign of each term within them. $$ - \left( a^2+2abi+b^2i^2 \right) = -a^2-2abi-b^2i^2 $$ |
⑥ | Combine like terms: $$ ac+bci+a^2+2abi+b^2i^2 = b^2i^2+2abi+bci+a^2+ac $$ |