Answer
$$\frac{2-3i}{(3+2i)\cdot(1-2i)-(2-3i)\cdot(3+3i)}=\frac{-1+2i}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-3i}{(3+2i)\cdot(1-2i)-(2-3i)\cdot(3+3i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2-3i}{3-6i+2i-4i^2-(6+6i-9i-9i^2)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2-3i}{-4i^2-4i+3-(-9i^2-3i+6)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{2-3i}{4-4i+3-(9-3i+6)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{2-3i}{-4i+7-(-3i+15)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{2-3i}{-4i+7+3i-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{2-3i}{-i-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} } }}}\frac{-1+2i}{5}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3+2i}\right) $ by each term in $ \left( 1-2i\right) $. $$ \left( \color{blue}{3+2i}\right) \cdot \left( 1-2i\right) = 3-6i+2i-4i^2 $$ |
| ② | Multiply each term of $ \left( \color{blue}{2-3i}\right) $ by each term in $ \left( 3+3i\right) $. $$ \left( \color{blue}{2-3i}\right) \cdot \left( 3+3i\right) = 6+6i-9i-9i^2 $$ |
| ③ | Combine like terms: $$ 3 \color{blue}{-6i} + \color{blue}{2i} -4i^2 = -4i^2 \color{blue}{-4i} +3 $$ |
| ④ | Combine like terms: $$ 6+ \color{blue}{6i} \color{blue}{-9i} -9i^2 = -9i^2 \color{blue}{-3i} +6 $$ |
| ⑤ | $$ -4i^2 = -4 \cdot (-1) = 4 $$ |
| ⑥ | $$ -9i^2 = -9 \cdot (-1) = 9 $$ |
| ⑦ | Combine like terms: $$ \color{blue}{4} -4i+ \color{blue}{3} = -4i+ \color{blue}{7} $$ |
| ⑧ | Combine like terms: $$ \color{blue}{9} -3i+ \color{blue}{6} = -3i+ \color{blue}{15} $$ |
| ⑨ | Remove the parentheses by changing the sign of each term within them. $$ - \left( -3i+15 \right) = 3i-15 $$ |
| ⑩ | Simplify denominator $$ \color{blue}{-4i} + \color{red}{7} + \color{blue}{3i} \color{red}{-15} = \color{blue}{-i} \color{red}{-8} $$ |
| ⑪ | Divide $ \, 2-3i \, $ by $ \, -8-i \, $ to get $\,\, \dfrac{-1+2i}{5} $. ( view steps ) |
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