Answer
$$-i(x+iy)=-i^2y-ix$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-i(x+iy)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-(1ix+i^2y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-ix-i^2y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-i^2y-ix\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{i} $ by $ \left( x+iy\right) $ $$ \color{blue}{i} \cdot \left( x+iy\right) = ix+i^2y $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left(ix+i^2y \right) = -ix-i^2y $$ |
| ③ | Combine like terms: $$ -i^2y-ix = -i^2y-ix $$ |
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