Question
Answer
$$(3-2i)\dfrac{2+3i}{(1+2i)\cdot(2-1)}=\dfrac{-19i+22}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2i)\frac{2+3i}{(1+2i)\cdot(2-1)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(3-2i)\frac{2+3i}{2-1+4i-2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(3-2i)\frac{2+3i}{2i+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(3-2i)\frac{8-i}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2i^2-19i+24}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-2-19i+24}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-19i+22}{5}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+2i}\right) $ by each term in $ \left( 2-1\right) $. $$ \left( \color{blue}{1+2i}\right) \cdot \left( 2-1\right) = 2-1+4i-2i $$ |
| ② | Combine like terms: $$ \color{blue}{2} \color{blue}{-1} + \color{red}{4i} \color{red}{-2i} = \color{red}{2i} + \color{blue}{1} $$ |
| ③ | Divide $ \, 2+3i \, $ by $ \, 1+2i \, $ to get $\,\, \dfrac{8-i}{5} $. ( view steps ) |
| ④ | Multiply $3-2i$ by $ \dfrac{8-i}{5} $ to get $ \dfrac{2i^2-19i+24}{5} $. 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. $$ \begin{aligned} 3-2i \cdot \frac{8-i}{5} & \xlongequal{\text{Step 1}} \frac{3-2i}{\color{red}{1}} \cdot \frac{8-i}{5} \xlongequal{\text{Step 2}} \frac{ \left( 3-2i \right) \cdot \left( 8-i \right) }{ 1 \cdot 5 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 24-3i-16i+2i^2 }{ 5 } = \frac{2i^2-19i+24}{5} \end{aligned} $$ |
| ⑤ | $$ 2i^2 = 2 \cdot (-1) = -2 $$ |
| ⑥ | Simplify numerator $$ \color{blue}{-2} -19i+ \color{blue}{24} = -19i+ \color{blue}{22} $$ |
This page was created using
Complex Expression calculator