User:Poketeacher/Sandbox

This is my personal sanbox

i
$$i = \sqrt{-1}$$

$$\sqrt{i} = \frac{1 + i}{\sqrt{2}}$$ because $$\left(\frac{1 + i}{\sqrt{2}}\right)^2 = \frac{(1 + i)(1 + i)}{2} = \frac{1 + i + i + i^2}{2} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$$

Algebra
$$\frac{2ab^2 \times 2bac^3}{3ab^2 \times b ac^4} = \frac{2a^22b^3c^3}{3a^2b^3c^4} = \frac{b^3}{a^2c}$$