Question
Answer
$$(1.95-2.44i)\dfrac{-4i}{5}=\dfrac{-8-4i}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1.95-2.44i)\frac{-4i}{5}& \xlongequal{ }(1.95-2i)\frac{-4i}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8i^2-4i}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-8-4i}{5}\end{aligned} $$ | |
| ① | Multiply $1-2i$ by $ \dfrac{-4i}{5} $ to get $ \dfrac{8i^2-4i}{5} $. Step 1: Write $ 1-2i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 1-2i \cdot \frac{-4i}{5} & \xlongequal{\text{Step 1}} \frac{1-2i}{\color{red}{1}} \cdot \frac{-4i}{5} \xlongequal{\text{Step 2}} \frac{ \left( 1-2i \right) \cdot \left( -4i \right) }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -4i+8i^2 }{ 5 } = \frac{8i^2-4i}{5} \end{aligned} $$ |
| ② | $$ 8i^2 = 8 \cdot (-1) = -8 $$ |
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