Question
Answer
$$z^3-(4+2i)z^2+(4+5i)z-(1+3i)=-2iz^2+z^3+5iz-4z^2-3i+4z-1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}z^3-(4+2i)z^2+(4+5i)z-(1+3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}z^3-(4z^2+2iz^2)+4z+5iz-(1+3i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}z^3-4z^2-2iz^2+4z+5iz-(1+3i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-2iz^2+z^3+5iz-4z^2+4z-(1+3i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-2iz^2+z^3+5iz-4z^2+4z-1-3i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-2iz^2+z^3+5iz-4z^2-3i+4z-1\end{aligned} $$ | |
| ① | $$ \left( \color{blue}{4+2i}\right) \cdot z^2 = 4z^2+2iz^2 $$$$ \left( \color{blue}{4+5i}\right) \cdot z = 4z+5iz $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4z^2+2iz^2 \right) = -4z^2-2iz^2 $$ |
| ③ | Combine like terms: $$ z^3-4z^2-2iz^2+4z+5iz = -2iz^2+z^3+5iz-4z^2+4z $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 1+3i \right) = -1-3i $$ |
| ⑤ | Combine like terms: $$ -2iz^2+z^3+5iz-4z^2-3i+4z-1 = -2iz^2+z^3+5iz-4z^2-3i+4z-1 $$ |
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