Question
Answer
$$\dfrac{1-i}{3+i}-\dfrac{3+2i}{1-2i}=\dfrac{-10i+2}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1-i}{3+i}-\frac{3+2i}{1-2i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1-2i}{5}-\frac{-1+8i}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-10i+2}{5}\end{aligned} $$ | |
| ① | Divide $ \, 1-i \, $ by $ \, 3+i \, $ to get $\,\, \dfrac{1-2i}{5} $. ( view steps )Divide $ \, 3+2i \, $ by $ \, 1-2i \, $ to get $\,\, \dfrac{-1+8i}{5} $. ( view steps ) |
| ② | Subtract $ \dfrac{-1+8i}{5} $ from $ \dfrac{1-2i}{5} $ to get $ \dfrac{-10i+2}{5} $. To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator. $$ \begin{aligned} \frac{1-2i}{5} - \frac{-1+8i}{5} & = \frac{1-2i}{\color{blue}{5}} - \frac{-1+8i}{\color{blue}{5}} =\frac{ 1-2i - \left( -1+8i \right) }{ \color{blue}{ 5 }} = \\[1ex] &= \frac{-10i+2}{5} \end{aligned} $$ |
This page was created using
Complex Expression calculator