Question
Answer
$$(3-2i)\dfrac{2+3i}{(1+2i)\cdot(2-i)}=\dfrac{-16i+63}{25}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2i)\frac{2+3i}{(1+2i)\cdot(2-i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(3-2i)\frac{2+3i}{2-i+4i-2i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(3-2i)\frac{2+3i}{-2i^2+3i+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(3-2i)\frac{2+3i}{2+3i+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(3-2i)\frac{2+3i}{3i+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}(3-2i)\frac{17+6i}{25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-12i^2-16i+51}{25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{12-16i+51}{25} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{-16i+63}{25}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+2i}\right) $ by each term in $ \left( 2-i\right) $. $$ \left( \color{blue}{1+2i}\right) \cdot \left( 2-i\right) = 2-i+4i-2i^2 $$ |
| ② | Combine like terms: $$ 2 \color{blue}{-i} + \color{blue}{4i} -2i^2 = -2i^2+ \color{blue}{3i} +2 $$ |
| ③ | $$ -2i^2 = -2 \cdot (-1) = 2 $$ |
| ④ | Combine like terms: $$ \color{blue}{2} +3i+ \color{blue}{2} = 3i+ \color{blue}{4} $$ |
| ⑤ | Divide $ \, 2+3i \, $ by $ \, 4+3i \, $ to get $\,\, \dfrac{17+6i}{25} $. ( view steps ) |
| ⑥ | Multiply $3-2i$ by $ \dfrac{17+6i}{25} $ to get $ \dfrac{-12i^2-16i+51}{25} $. Step 1: Write $ 3-2i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 3-2i \cdot \frac{17+6i}{25} & \xlongequal{\text{Step 1}} \frac{3-2i}{\color{red}{1}} \cdot \frac{17+6i}{25} \xlongequal{\text{Step 2}} \frac{ \left( 3-2i \right) \cdot \left( 17+6i \right) }{ 1 \cdot 25 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 51+18i-34i-12i^2 }{ 25 } = \frac{-12i^2-16i+51}{25} \end{aligned} $$ |
| ⑦ | $$ -12i^2 = -12 \cdot (-1) = 12 $$ |
| ⑧ | Simplify numerator $$ \color{blue}{12} -16i+ \color{blue}{51} = -16i+ \color{blue}{63} $$ |
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