Question
Answer
$$(3-2i)\dfrac{1+2i}{2-i}\cdot(2+3i)=9i^2+12i+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-2i)\frac{1+2i}{2-i}\cdot(2+3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(3-2i)\cdot1i\cdot(2+3i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(3i-2i^2)\cdot(2+3i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(3i+2)\cdot(2+3i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}6i+9i^2+4+6i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}9i^2+12i+4\end{aligned} $$ | |
| ① | Divide $ \, 1+2i \, $ by $ \, 2-i \, $ to get $\,\, i $. ( view steps ) |
| ② | $$ \left( \color{blue}{3-2i}\right) \cdot i = 3i-2i^2 $$ |
| ③ | $$ -2i^2 = -2 \cdot (-1) = 2 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{3i+2}\right) $ by each term in $ \left( 2+3i\right) $. $$ \left( \color{blue}{3i+2}\right) \cdot \left( 2+3i\right) = 6i+9i^2+4+6i $$ |
| ⑤ | Combine like terms: $$ \color{blue}{6i} +9i^2+4+ \color{blue}{6i} = 9i^2+ \color{blue}{12i} +4 $$ |
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