Answer
$$(1-6zi)\cdot(4-i)+(1+zi)\cdot(1+i)=7i^2z-23iz+5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-6zi)\cdot(4-i)+(1+zi)\cdot(1+i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4-i-24iz+6i^2z+1+i+iz+i^2z \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}7i^2z-23iz+5\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1-6iz}\right) $ by each term in $ \left( 4-i\right) $. $$ \left( \color{blue}{1-6iz}\right) \cdot \left( 4-i\right) = 4-i-24iz+6i^2z $$Multiply each term of $ \left( \color{blue}{1+iz}\right) $ by each term in $ \left( 1+i\right) $. $$ \left( \color{blue}{1+iz}\right) \cdot \left( 1+i\right) = 1+i+iz+i^2z $$ |
| ② | Combine like terms: $$ \color{blue}{4} \, \color{red}{ -\cancel{i}} \, \color{orange}{-24iz} + \color{blue}{6i^2z} + \color{blue}{1} + \, \color{red}{ \cancel{i}} \,+ \color{orange}{iz} + \color{blue}{i^2z} = \color{blue}{7i^2z} \color{orange}{-23iz} + \color{blue}{5} $$ |
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