Question
Answer
$$-\dfrac{2}{(3-ia)^2+4}=\dfrac{2}{a^2i^2-6ai+13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{2}{(3-ia)^2+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{9-6ai+a^2i^2+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2}{a^2i^2-6ai+13}\end{aligned} $$ | |
| ① | Find $ \left(3-ai\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}{ ai }$. $$ \begin{aligned}\left(3-ai\right)^2 = \color{blue}{3^2} -2 \cdot 3 \cdot ai + \color{red}{\left( ai \right)^2} = 9-6ai+a^2i^2\end{aligned} $$ |
| ② | Simplify denominator $$ \color{blue}{9} -6ai+a^2i^2+ \color{blue}{4} = a^2i^2-6ai+ \color{blue}{13} $$ |
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