Question
Answer
$$2pi\dfrac{i}{ln\cdot8+2pii}=\dfrac{2i^2p}{2i^2p+8ln}$$
Explanation
| $$ \begin{aligned}2pi\frac{i}{ln\cdot8+2pii}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2i^2p}{2i^2p+8ln}\end{aligned} $$ | |
| ① | Multiply $2ip$ by $ \dfrac{i}{8ln+2i^2p} $ to get $ \dfrac{2i^2p}{2i^2p+8ln} $. Step 1: Write $ 2ip $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2ip \cdot \frac{i}{8ln+2i^2p} & \xlongequal{\text{Step 1}} \frac{2ip}{\color{red}{1}} \cdot \frac{i}{8ln+2i^2p} \xlongequal{\text{Step 2}} \frac{ 2ip \cdot i }{ 1 \cdot \left( 8ln+2i^2p \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2i^2p }{ 8ln+2i^2p } = \frac{2i^2p}{2i^2p+8ln} \end{aligned} $$ |
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