Question
Answer
$$(3-2yi)\cdot(5+3i)-(3-2yi)\cdot(5+4i)=2i^2y-3i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2yi)\cdot(5+3i)-(3-2yi)\cdot(5+4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}15+9i-10iy-6i^2y-(15+12i-10iy-8i^2y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}15+9i-10iy-6i^2y-15-12i+10iy+8i^2y \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{15}+9i -\cancel{10iy}-6i^2y -\cancel{15}-12i+ \cancel{10iy}+8i^2y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2i^2y-3i\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-2iy}\right) $ by each term in $ \left( 5+3i\right) $. $$ \left( \color{blue}{3-2iy}\right) \cdot \left( 5+3i\right) = 15+9i-10iy-6i^2y $$Multiply each term of $ \left( \color{blue}{3-2iy}\right) $ by each term in $ \left( 5+4i\right) $. $$ \left( \color{blue}{3-2iy}\right) \cdot \left( 5+4i\right) = 15+12i-10iy-8i^2y $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 15+12i-10iy-8i^2y \right) = -15-12i+10iy+8i^2y $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{15}} \,+ \color{green}{9i} \, \color{orange}{ -\cancel{10iy}} \, \color{red}{-6i^2y} \, \color{blue}{ -\cancel{15}} \, \color{green}{-12i} + \, \color{orange}{ \cancel{10iy}} \,+ \color{red}{8i^2y} = \color{red}{2i^2y} \color{green}{-3i} $$ |
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