Question
Answer
$$(2+3i)\dfrac{2+i}{1}-i=7i+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2+3i)\frac{2+i}{1}-i& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3i^2+8i+4-i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-3+8i+4-i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}7i+1\end{aligned} $$ | |
| ① | Multiply $2+3i$ by $ \dfrac{2+i}{1} $ to get $ 4+2i+6i+3i^2$. Step 1: Write $ 2+3i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2+3i \cdot \frac{2+i}{1} & \xlongequal{\text{Step 1}} \frac{2+3i}{\color{red}{1}} \cdot \frac{2+i}{1} \xlongequal{\text{Step 2}} \frac{ \left( 2+3i \right) \cdot \left( 2+i \right) }{ 1 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4+2i+6i+3i^2 }{ 1 } = \frac{3i^2+8i+4}{1} =4+2i+6i+3i^2 \end{aligned} $$ |
| ② | $$ 3i^2 = 3 \cdot (-1) = -3 $$ |
| ③ | Combine like terms: $$ \color{blue}{-3} + \color{red}{8i} + \color{blue}{4} \color{red}{-i} = \color{red}{7i} + \color{blue}{1} $$ |
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