Answer
$$(3-5zi)\cdot(5+2i)-(3-4zi)\cdot(5+2i)=-2i^2z-5iz$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-5zi)\cdot(5+2i)-(3-4zi)\cdot(5+2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}15+6i-25iz-10i^2z-(15+6i-20iz-8i^2z) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}15+6i-25iz-10i^2z-15-6i+20iz+8i^2z \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{15}+ \cancel{6i}-25iz-10i^2z -\cancel{15} -\cancel{6i}+20iz+8i^2z \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-2i^2z-5iz\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3-5iz}\right) $ by each term in $ \left( 5+2i\right) $. $$ \left( \color{blue}{3-5iz}\right) \cdot \left( 5+2i\right) = 15+6i-25iz-10i^2z $$Multiply each term of $ \left( \color{blue}{3-4iz}\right) $ by each term in $ \left( 5+2i\right) $. $$ \left( \color{blue}{3-4iz}\right) \cdot \left( 5+2i\right) = 15+6i-20iz-8i^2z $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 15+6i-20iz-8i^2z \right) = -15-6i+20iz+8i^2z $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{15}} \,+ \, \color{green}{ \cancel{6i}} \, \color{blue}{-25iz} \color{red}{-10i^2z} \, \color{blue}{ -\cancel{15}} \, \, \color{green}{ -\cancel{6i}} \,+ \color{blue}{20iz} + \color{red}{8i^2z} = \color{red}{-2i^2z} \color{blue}{-5iz} $$ |
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