Question
Answer
$$(5+6i)\cdot6\cdot\dfrac{i}{(3-2i)\cdot(4+2i)}=\dfrac{204i-318}{130}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(5+6i)\cdot6 \cdot \frac{i}{(3-2i)\cdot(4+2i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(30+36i)\frac{i}{(3-2i)\cdot(4+2i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(30+36i)\frac{i}{12+6i-8i-4i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(30+36i)\frac{i}{-4i^2-2i+12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(30+36i)\frac{i}{4-2i+12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}(30+36i)\frac{i}{-2i+16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}(30+36i)\frac{-1+8i}{130} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{288i^2+204i-30}{130} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{-288+204i-30}{130} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{204i-318}{130}\end{aligned} $$ | |
| ① | $$ \left( \color{blue}{5+6i}\right) \cdot 6 = 30+36i $$ |
| ② | Multiply each term of $ \left( \color{blue}{3-2i}\right) $ by each term in $ \left( 4+2i\right) $. $$ \left( \color{blue}{3-2i}\right) \cdot \left( 4+2i\right) = 12+6i-8i-4i^2 $$ |
| ③ | Combine like terms: $$ 12+ \color{blue}{6i} \color{blue}{-8i} -4i^2 = -4i^2 \color{blue}{-2i} +12 $$ |
| ④ | $$ -4i^2 = -4 \cdot (-1) = 4 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{4} -2i+ \color{blue}{12} = -2i+ \color{blue}{16} $$ |
| ⑥ | Divide $ \, i \, $ by $ \, 16-2i \, $ to get $\,\, \dfrac{-1+8i}{130} $. ( view steps ) |
| ⑦ | Multiply $30+36i$ by $ \dfrac{-1+8i}{130} $ to get $ \dfrac{288i^2+204i-30}{130} $. Step 1: Write $ 30+36i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 30+36i \cdot \frac{-1+8i}{130} & \xlongequal{\text{Step 1}} \frac{30+36i}{\color{red}{1}} \cdot \frac{-1+8i}{130} \xlongequal{\text{Step 2}} \frac{ \left( 30+36i \right) \cdot \left( -1+8i \right) }{ 1 \cdot 130 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -30+240i-36i+288i^2 }{ 130 } = \frac{288i^2+204i-30}{130} \end{aligned} $$ |
| ⑧ | $$ 288i^2 = 288 \cdot (-1) = -288 $$ |
| ⑨ | Simplify numerator $$ \color{blue}{-288} +204i \color{blue}{-30} = 204i \color{blue}{-318} $$ |
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