Question
Answer
$$\dfrac{(1-4i)\cdot(3+i)}{1+3i}=\dfrac{-13-16i}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(1-4i)\cdot(3+i)}{1+3i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+i-12i-4i^2}{1+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3+i-12i+4}{1+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-11i+7}{1+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-13-16i}{5}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1-4i}\right) $ by each term in $ \left( 3+i\right) $. $$ \left( \color{blue}{1-4i}\right) \cdot \left( 3+i\right) = 3+i-12i-4i^2 $$ |
| ② | $$ -4i^2 = -4 \cdot (-1) = 4 $$ |
| ③ | Simplify numerator $$ \color{blue}{3} + \color{red}{i} \color{red}{-12i} + \color{blue}{4} = \color{red}{-11i} + \color{blue}{7} $$ |
| ④ | Divide $ \, 7-11i \, $ by $ \, 1+3i \, $ to get $\,\, \dfrac{-13-16i}{5} $. ( view steps ) |
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