Answer
$$\frac{1}{(3+i)^2}=\frac{1}{6i+8}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{(3+i)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{9+6i+i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{9+6i-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{6i+8}\end{aligned} $$ | |
| 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 }$. $$ \begin{aligned}\left(3+i\right)^2 = \color{blue}{3^2} +2 \cdot 3 \cdot i + \color{red}{i^2} = 9+6i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Simplify denominator $$ \color{blue}{9} +6i \color{blue}{-1} = 6i+ \color{blue}{8} $$ |
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