Question
Answer
$$(5-3yi)\cdot(3+4i)+(1+3yi)\cdot(3+4i)=24i+18$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(5-3yi)\cdot(3+4i)+(1+3yi)\cdot(3+4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}15+20i-9iy-12i^2y+3+4i+9iy+12i^2y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}24i+18\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{5-3iy}\right) $ by each term in $ \left( 3+4i\right) $. $$ \left( \color{blue}{5-3iy}\right) \cdot \left( 3+4i\right) = 15+20i-9iy-12i^2y $$Multiply each term of $ \left( \color{blue}{1+3iy}\right) $ by each term in $ \left( 3+4i\right) $. $$ \left( \color{blue}{1+3iy}\right) \cdot \left( 3+4i\right) = 3+4i+9iy+12i^2y $$ |
| ② | Combine like terms: $$ \color{blue}{15} + \color{red}{20i} \, \color{green}{ -\cancel{9iy}} \, \, \color{blue}{ -\cancel{12i^2y}} \,+ \color{blue}{3} + \color{red}{4i} + \, \color{green}{ \cancel{9iy}} \,+ \, \color{blue}{ \cancel{12i^2y}} \, = \color{red}{24i} + \color{blue}{18} $$ |
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