Answer
$$\frac{(2+3i)^2}{2-i}=\frac{-22+19i}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(2+3i)^2}{2-i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+12i+9i^2}{2-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4+12i-9}{2-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12i-5}{2-i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-22+19i}{5}\end{aligned} $$ | |
| ① | Find $ \left(2+3i\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2 } $ and $ B = \color{red}{ 3i }$. $$ \begin{aligned}\left(2+3i\right)^2 = \color{blue}{2^2} +2 \cdot 2 \cdot 3i + \color{red}{\left( 3i \right)^2} = 4+12i+9i^2\end{aligned} $$ |
| ② | $$ 9i^2 = 9 \cdot (-1) = -9 $$ |
| ③ | Simplify numerator $$ \color{blue}{4} +12i \color{blue}{-9} = 12i \color{blue}{-5} $$ |
| ④ | Divide $ \, -5+12i \, $ by $ \, 2-i \, $ to get $\,\, \dfrac{-22+19i}{5} $. ( view steps ) |
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