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