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Question

Answer

$$\dfrac{1}{1-8i}+\dfrac{1}{1+8i}=\dfrac{2}{65}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{1}{1-8i}+\frac{1}{1+8i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{-64i^2+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2}{64+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2}{65}\end{aligned} $$
①

Add $ \dfrac{1}{1-8i} $ and $ \dfrac{1}{1+8i} $ to get $ \dfrac{ \color{purple}{ 2 } }{ -64i^2+1 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 8i+1 }$ and the second by $\color{blue}{ -8i+1 }$.

$$ \begin{aligned} \frac{1}{1-8i} + \frac{1}{1+8i} & = \frac{ 1 \cdot \color{blue}{ \left( 8i+1 \right) }}{ \left( 1-8i \right) \cdot \color{blue}{ \left( 8i+1 \right) }} + \frac{ 1 \cdot \color{blue}{ \left( -8i+1 \right) }}{ \left( 1+8i \right) \cdot \color{blue}{ \left( -8i+1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 8i+1 } }{ \cancel{8i}+1-64i^2 -\cancel{8i} } + \frac{ \color{purple}{ -8i+1 } }{ \cancel{8i}+1-64i^2 -\cancel{8i} } = \\[1ex] &=\frac{ \color{purple}{ 2 } }{ -64i^2+1 } \end{aligned} $$
②
$$ -64i^2 = -64 \cdot (-1) = 64 $$
③

Simplify denominator

$$ \color{blue}{64} + \color{blue}{1} = \color{blue}{65} $$
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