Question
Answer
$$\dfrac{1}{-w(-w+i)}=\dfrac{1}{w^2-iw}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{-w(-w+i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{-(-w^2+iw)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{w^2-iw}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{w} $ by $ \left( -w+i\right) $ $$ \color{blue}{w} \cdot \left( -w+i\right) = -w^2+iw $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(-w^2+iw \right) = w^2-iw $$ |
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