Question
Answer
$$\dfrac{4i}{(1+i)\cdot(2+i)\cdot(3+i)}=\dfrac{2}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4i}{(1+i)\cdot(2+i)\cdot(3+i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4i}{(2+i+2i+i^2)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4i}{(1i^2+3i+2)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4i}{(-1+3i+2)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{4i}{(3i+1)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{4i}{9i+3i^2+3+i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{4i}{9i-3+3+i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{4i}{10i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{2}{5}\end{aligned} $$ | |
| 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 $ \, 4i \, $ by $ \, 10i \, $ to get $\,\, \dfrac{2}{5} $. ( view steps ) |
This page was created using
Complex Expression calculator