Question
Answer
$$\dfrac{2\cdot(-1+i)(z+i-1)-(z-2)i}{(z-1+i)\cdot(-1+i)-(z-2)i}=\dfrac{2i^2+iz-2i-2z+2}{i^2-z+1}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\cdot(-1+i)(z+i-1)-(z-2)i}{(z-1+i)\cdot(-1+i)-(z-2)i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{(-2+2i)(z+i-1)-(z-2)i}{-z+iz+1-i-i+i^2-(1iz-2i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-2z-2i+2+2iz+2i^2-2i-(1iz-2i)}{i^2+iz-2i-z+1-(1iz-2i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{2i^2+2iz-4i-2z+2-(1iz-2i)}{i^2+iz-2i-z+1-(1iz-2i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{2i^2+2iz-4i-2z+2-iz+2i}{i^2+iz-2i-z+1-iz+2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} } }}}\frac{2i^2+iz-2i-2z+2}{i^2+iz-2i-z+1-iz+2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{2i^2+iz-2i-2z+2}{i^2-z+1}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2} $ by $ \left( -1+i\right) $ $$ \color{blue}{2} \cdot \left( -1+i\right) = -2+2i $$ |
| ② | Multiply each term of $ \left( \color{blue}{z-1+i}\right) $ by each term in $ \left( -1+i\right) $. $$ \left( \color{blue}{z-1+i}\right) \cdot \left( -1+i\right) = -z+iz+1-i-i+i^2 $$ |
| ③ | $$ \left( \color{blue}{z-2}\right) \cdot i = iz-2i $$ |
| ④ | Multiply each term of $ \left( \color{blue}{-2+2i}\right) $ by each term in $ \left( z+i-1\right) $. $$ \left( \color{blue}{-2+2i}\right) \cdot \left( z+i-1\right) = -2z-2i+2+2iz+2i^2-2i $$ |
| ⑤ | $$ \left( \color{blue}{z-2}\right) \cdot i = iz-2i $$ |
| ⑥ | Combine like terms: $$ -z+iz+1 \color{blue}{-i} \color{blue}{-i} +i^2 = i^2+iz \color{blue}{-2i} -z+1 $$ |
| ⑦ | Combine like terms: $$ -2z \color{blue}{-2i} +2+2iz+2i^2 \color{blue}{-2i} = 2i^2+2iz \color{blue}{-4i} -2z+2 $$ |
| ⑧ | Combine like terms: $$ -z+iz+1 \color{blue}{-i} \color{blue}{-i} +i^2 = i^2+iz \color{blue}{-2i} -z+1 $$ |
| ⑨ | Remove the parentheses by changing the sign of each term within them. $$ - \left( iz-2i \right) = -iz+2i $$ |
| ⑩ | Remove the parentheses by changing the sign of each term within them. $$ - \left( iz-2i \right) = -iz+2i $$ |
| ⑪ | Simplify numerator $$ 2i^2+ \color{blue}{2iz} \color{red}{-4i} -2z+2 \color{blue}{-iz} + \color{red}{2i} = 2i^2+ \color{blue}{iz} \color{red}{-2i} -2z+2 $$ |
| ⑫ | Simplify denominator $$ i^2+ \, \color{blue}{ \cancel{iz}} \, \, \color{green}{ -\cancel{2i}} \,-z+1 \, \color{blue}{ -\cancel{iz}} \,+ \, \color{green}{ \cancel{2i}} \, = i^2-z+1 $$ |
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