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