$$1=1$$ $$\Rightarrow\frac{-1}{1}=\frac{1}{-1}$$ $$\Rightarrow \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$ $$\Rightarrow\frac{i}{1}=\frac{1}{i}$$ $$\Rightarrow\frac{i}{2}=\frac{1}{2i}$$ $$\Rightarrow\frac{i}{2}+\frac{3}{2i} = \frac{1}{2i} +\frac{3}{2i}$$ $$\Rightarrow i(\frac{i}{2}+\frac{3}{2i} ) = i(\frac{1}{2i} +\frac{3}{2i})$$ $$\Rightarrow\frac{-1}{2}=\frac{1}{2}$$ $$\Rightarrow1=2$$
What is wrong in this?
$$\sqrt{\frac{a}{b}} \neq \frac{\sqrt{a}}{\sqrt{b}} $$ in general unless both $a$ and $b$ are positive.
The trick is assuming that $$\sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1}}\Rightarrow \frac{\sqrt{-1}}{\sqrt{1}} = \frac{\sqrt{1}}{\sqrt{-1}}$$ which is not true because by definition of (complex) square root $-1=\sqrt{-1}\cdot \sqrt{-1} \not=\sqrt{1}\cdot \sqrt{1}=1$.
On the other hand, note that if $a,b,c,d$ are real POSITIVE numbers then $$\sqrt{\frac{a}{b}}=\sqrt{\frac{c}{d}}\Leftrightarrow \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{c}}{\sqrt{d}}.$$
Let's take a closer look at the general theory, the complex plane.
Take the complex plane where z=r+ic, i beeing a square root of -1. Note that -i is the other number that squares to 1.
In polar coordinates z=r e^(i phi). You can add any number of 2 pi radians (360 deg, full circles) to phi and make it phi +2 pi k and the point is the same. When you take n:th root, you must take a normal real root of r but also divide the angle (phi + 2k pi) with n and thus you get n different n:th roots with k=0,1,2,...n-1. With greater k they repeat. Most of them are off the x axis or, off the real line. So, actually there are n complex numbers distributed evenly around the circle, that give a real number r (e.g. +1) when raised to power n. This you have forgotten in the proof. Read "roots of unity" in Wikipedia.
For 1 you have phi=0 deg, and multiples of 360 degrees ( 2 pi radians) because 1 remains on the positive x axis no matter how many times you go around the origo. There are two numbers, with 0 and pi radians then, that square to 1.
-1 has 180 deg or pi radians as angle, and thus the square root has pi/2 +k pi, or i at the top and -i at the bottom of the circle.
This all explains why roots of negative numbers do not obey all the familiar laws of real number arithmetic. If you assume that there is only one number that squares to 1 (as you have done with the roots equality) then you kind of assume that 1=-1 and this leads to funny results.
