Question
Answer
$$\dfrac{(-2+i)(x+iy)-2i}{-1(x+iy)}=\dfrac{-2x-2iy+ix+i^2y-2i}{-x-iy}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(-2+i)(x+iy)-2i}{-1(x+iy)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-2x-2iy+ix+i^2y-2i}{-1(x+iy)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-2x-2iy+ix+i^2y-2i}{-x-iy}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{-2+i}\right) $ by each term in $ \left( x+iy\right) $. $$ \left( \color{blue}{-2+i}\right) \cdot \left( x+iy\right) = -2x-2iy+ix+i^2y $$ |
| ② | Multiply $ \color{blue}{-1} $ by $ \left( x+iy\right) $ $$ \color{blue}{-1} \cdot \left( x+iy\right) = -x-iy $$ |
This page was created using
Complex Expression calculator