Mathematical Fallacy: Is $1=2$?

$\begin{aligned} \dfrac{-1}{1}=&\dfrac{1}{-1} \quad & \ldots(1) \\ \sqrt{\dfrac{-1}{1}}=&\sqrt{\dfrac{1}{-1}} \quad & \ldots (2) \\ \dfrac{\sqrt{-1}}{\sqrt{1}}=& \dfrac{\sqrt{1}}{\sqrt{-1}} \quad &\ldots (3) \\ \dfrac{i}{1}=&\dfrac{1}{i} \quad & \ldots(4) \\ \dfrac{i}{2}=&\dfrac{1}{2i} \quad & \ldots(5) \\ \dfrac{i}{2}+\dfrac{3}{2i}=&\dfrac{1}{2i}+\dfrac{3}{2i} \quad &\ldots(6) \\ i\cdot\left(\dfrac{i}{2}+\dfrac{3}{2i}\right)=& i \cdot\left( \dfrac{1}{2i}+\dfrac{3}{2i}\right) \quad & \ldots(7) \\ \dfrac{i^2}{2}+\dfrac{3i}{2i}=& \dfrac{i}{2i}+\dfrac{3i}{2i} \quad & \ldots(8) \\ \dfrac{-1}{2}+\dfrac{3}{2}=& \dfrac{1}{2}+\dfrac{3}{2} \quad & \ldots (9) \\ 1=&2 \end{aligned}$

Consider the steps above.

In which step is the mistake committed?

Note: $i=\sqrt{-1}$ in equation $(9)$ and the last one we assume to work out the L.H.S. and the R.H.S. separately.

Source:Spencer,1998