Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(1+i)\cdot(1-2i)}{(2+i)\cdot(4-3i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1-2i+i-2i^2}{8-6i+4i-3i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1-2i+i+2}{8-6i+4i-3i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1-2i+i+2}{8-6i+4i+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-i+3}{8-6i+4i+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-i+3}{-2i+11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{7-i}{25}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+i}\right) $ by each term in $ \left( 1-2i\right) $. $$ \left( \color{blue}{1+i}\right) \cdot \left( 1-2i\right) = 1-2i+i-2i^2 $$ |
| ② | Multiply each term of $ \left( \color{blue}{2+i}\right) $ by each term in $ \left( 4-3i\right) $. $$ \left( \color{blue}{2+i}\right) \cdot \left( 4-3i\right) = 8-6i+4i-3i^2 $$ |
| ③ | $$ -2i^2 = -2 \cdot (-1) = 2 $$ |
| ④ | $$ -3i^2 = -3 \cdot (-1) = 3 $$ |
| ⑤ | Simplify numerator $$ \color{blue}{1} \color{red}{-2i} + \color{red}{i} + \color{blue}{2} = \color{red}{-i} + \color{blue}{3} $$ |
| ⑥ | Simplify denominator $$ \color{blue}{8} \color{red}{-6i} + \color{red}{4i} + \color{blue}{3} = \color{red}{-2i} + \color{blue}{11} $$ |
| ⑦ | Divide $ \, 3-i \, $ by $ \, 11-2i \, $ to get $\,\, \dfrac{7-i}{25} $. ( view steps ) |