Question
Answer
$$\dfrac{2+3i+1-4i}{(8-2i)\cdot(1-i)}=\dfrac{7+6i}{34}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2+3i+1-4i}{(8-2i)\cdot(1-i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-i+3}{8-8i-2i+2i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-i+3}{8-8i-2i-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-i+3}{-10i+6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{7+6i}{34}\end{aligned} $$ | |
| ① | Combine like terms: $$ \color{blue}{2} + \color{red}{3i} + \color{blue}{1} \color{red}{-4i} = \color{red}{-i} + \color{blue}{3} $$ |
| ② | Multiply each term of $ \left( \color{blue}{8-2i}\right) $ by each term in $ \left( 1-i\right) $. $$ \left( \color{blue}{8-2i}\right) \cdot \left( 1-i\right) = 8-8i-2i+2i^2 $$ |
| ③ | $$ 2i^2 = 2 \cdot (-1) = -2 $$ |
| ④ | Simplify denominator $$ \color{blue}{8} \color{red}{-8i} \color{red}{-2i} \color{blue}{-2} = \color{red}{-10i} + \color{blue}{6} $$ |
| ⑤ | Divide $ \, 3-i \, $ by $ \, 6-10i \, $ to get $\,\, \dfrac{7+6i}{34} $. ( view steps ) |
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