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