Question
Answer
$$\dfrac{2i}{(1-i)\cdot(2-i)\cdot(3-i)}=-\dfrac{1}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2i}{(1-i)\cdot(2-i)\cdot(3-i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2i}{(2-i-2i+i^2)\cdot(3-i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2i}{(1i^2-3i+2)\cdot(3-i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2i}{(-1-3i+2)\cdot(3-i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2i}{(-3i+1)\cdot(3-i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{2i}{-9i+3i^2+3-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{2i}{-9i-3+3-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{2i}{-10i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}-\frac{1}{5}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1-i}\right) $ by each term in $ \left( 2-i\right) $. $$ \left( \color{blue}{1-i}\right) \cdot \left( 2-i\right) = 2-i-2i+i^2 $$ |
| ② | Combine like terms: $$ 2 \color{blue}{-i} \color{blue}{-2i} +i^2 = i^2 \color{blue}{-3i} +2 $$ |
| ③ | $$ i^2 = -1 $$ |
| ④ | Combine like terms: $$ \color{blue}{-1} -3i+ \color{blue}{2} = -3i+ \color{blue}{1} $$ |
| ⑤ | Multiply each term of $ \left( \color{blue}{-3i+1}\right) $ by each term in $ \left( 3-i\right) $. $$ \left( \color{blue}{-3i+1}\right) \cdot \left( 3-i\right) = -9i+3i^2+3-i $$ |
| ⑥ | $$ 3i^2 = 3 \cdot (-1) = -3 $$ |
| ⑦ | Simplify denominator $$ \color{blue}{-9i} \, \color{red}{ -\cancel{3}} \,+ \, \color{red}{ \cancel{3}} \, \color{blue}{-i} = \color{blue}{-10i} $$ |
| ⑧ | Divide $ \, 2i \, $ by $ \, -10i \, $ to get $\,\, \dfrac{-1}{5} $. ( view steps ) |
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