Question
Answer
$$\dfrac{2c+i(a^2+b^2+c^2-1)}{2a+2ib}=\dfrac{a^2i+b^2i+c^2i+2c-i}{2a+2ib}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2c+i(a^2+b^2+c^2-1)}{2a+2ib}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2c+a^2i+b^2i+c^2i-i}{2a+2ib} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{a^2i+b^2i+c^2i+2c-i}{2a+2ib}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{i} $ by $ \left( a^2+b^2+c^2-1\right) $ $$ \color{blue}{i} \cdot \left( a^2+b^2+c^2-1\right) = a^2i+b^2i+c^2i-i $$ |
| ② | Combine like terms: $$ 2c+a^2i+b^2i+c^2i-i = a^2i+b^2i+c^2i+2c-i $$ |
This page was created using
Complex Expression calculator