Answer
$$(1-2zi)\cdot(2+5i)-(1-2zi)\cdot(2-3i)=-16i^2z+8i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-2zi)\cdot(2+5i)-(1-2zi)\cdot(2-3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2+5i-4iz-10i^2z-(2-3i-4iz+6i^2z) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2+5i-4iz-10i^2z-2+3i+4iz-6i^2z \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{2}+5i -\cancel{4iz}-10i^2z -\cancel{2}+3i+ \cancel{4iz}-6i^2z \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-16i^2z+8i\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1-2iz}\right) $ by each term in $ \left( 2+5i\right) $. $$ \left( \color{blue}{1-2iz}\right) \cdot \left( 2+5i\right) = 2+5i-4iz-10i^2z $$Multiply each term of $ \left( \color{blue}{1-2iz}\right) $ by each term in $ \left( 2-3i\right) $. $$ \left( \color{blue}{1-2iz}\right) \cdot \left( 2-3i\right) = 2-3i-4iz+6i^2z $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 2-3i-4iz+6i^2z \right) = -2+3i+4iz-6i^2z $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{2}} \,+ \color{green}{5i} \, \color{orange}{ -\cancel{4iz}} \, \color{red}{-10i^2z} \, \color{blue}{ -\cancel{2}} \,+ \color{green}{3i} + \, \color{orange}{ \cancel{4iz}} \, \color{red}{-6i^2z} = \color{red}{-16i^2z} + \color{green}{8i} $$ |
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