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