Answer
$$(3-i)^2+(1-4i)^2=-14i-7$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3-i)^2+(1-4i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9-6i+i^2+1-8i+16i^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9-6i-1+1-8i-16 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-6i+8-8i-15 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-14i-7\end{aligned} $$ | |
| ① | Find $ \left(3-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}{ 3 } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(3-i\right)^2 = \color{blue}{3^2} -2 \cdot 3 \cdot i + \color{red}{i^2} = 9-6i+i^2\end{aligned} $$Find $ \left(1-4i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ 4i }$. $$ \begin{aligned}\left(1-4i\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot 4i + \color{red}{\left( 4i \right)^2} = 1-8i+16i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$$$ 16i^2 = 16 \cdot (-1) = -16 $$ |
| ③ | Combine like terms: $$ \color{blue}{9} -6i \color{blue}{-1} = -6i+ \color{blue}{8} $$Combine like terms: $$ \color{blue}{1} -8i \color{blue}{-16} = -8i \color{blue}{-15} $$ |
| ④ | Combine like terms: $$ \color{blue}{-6i} + \color{red}{8} \color{blue}{-8i} \color{red}{-15} = \color{blue}{-14i} \color{red}{-7} $$ |
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