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