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Question

Answer

$$\dfrac{1}{i\cdot4}-\dfrac{1}{1+i\cdot3.8}=\dfrac{-2+i}{20}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{1}{i\cdot4}-\frac{1}{1+i\cdot3.8}& \xlongequal{ }\frac{1}{i\cdot4}-\frac{1}{1+3i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-i+1}{12i^2+4i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-i+1}{-12+4i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-2+i}{20}\end{aligned} $$
①

Subtract $ \dfrac{1}{1+3i} $ from $ \dfrac{1}{4i} $ to get $ \dfrac{ \color{purple}{ -i+1 } }{ 12i^2+4i }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 3i+1 }$ and the second by $\color{blue}{ 4i }$.

$$ \begin{aligned} \frac{1}{4i} - \frac{1}{1+3i} & = \frac{ 1 \cdot \color{blue}{ \left( 3i+1 \right) }}{ 4i \cdot \color{blue}{ \left( 3i+1 \right) }} - \frac{ 1 \cdot \color{blue}{ 4i }}{ \left( 1+3i \right) \cdot \color{blue}{ 4i }} = \\[1ex] &=\frac{ \color{purple}{ 3i+1 } }{ 12i^2+4i } - \frac{ \color{purple}{ 4i } }{ 12i^2+4i }=\frac{ \color{purple}{ -i+1 } }{ 12i^2+4i } \end{aligned} $$
②
$$ 12i^2 = 12 \cdot (-1) = -12 $$
③
Divide $ \, 1-i \, $ by $ \, -12+4i \, $ to get $\,\, \dfrac{-2+i}{20} $.
( view steps )
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