Question
Answer
$$i\cdot(2-3i)-(5-i)^2=12i-21$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}i\cdot(2-3i)-(5-i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}i\cdot(2-3i)-(25-10i+i^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}i\cdot(2-3i)-(25-10i-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}i\cdot(2-3i)-(-10i+24) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2i-3i^2-(-10i+24) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2i+3-(-10i+24) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}2i+3+10i-24 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}12i-21\end{aligned} $$ | |
| ① | Find $ \left(5-i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 5 } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(5-i\right)^2 = \color{blue}{5^2} -2 \cdot 5 \cdot i + \color{red}{i^2} = 25-10i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \color{blue}{25} -10i \color{blue}{-1} = -10i+ \color{blue}{24} $$ |
| ④ | Multiply $ \color{blue}{i} $ by $ \left( 2-3i\right) $ $$ \color{blue}{i} \cdot \left( 2-3i\right) = 2i-3i^2 $$ |
| ⑤ | $$ -3i^2 = -3 \cdot (-1) = 3 $$ |
| ⑥ | Remove the parentheses by changing the sign of each term within them. $$ - \left( -10i+24 \right) = 10i-24 $$ |
| ⑦ | Combine like terms: $$ \color{blue}{2i} + \color{red}{3} + \color{blue}{10i} \color{red}{-24} = \color{blue}{12i} \color{red}{-21} $$ |
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