Simpler than undetermined coefficients is the following rule I discovered as a teenager.
Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $
and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$
Here $\:1+\sqrt{3}/2\:$ has norm $= 1/4.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 1/2\ $ yields $\ 1/2+\sqrt{3}/2\:$
and this has $\rm\ \sqrt{trace}\: =\: 1,\ \ thus,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1/2+\sqrt{3}/2.$
Below is another example.
Note $\:9-4\sqrt{2}\:$ has norm $= 49.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 7\ $ yields $\ 2-4\sqrt{2}\:$
and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ so,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1-2\sqrt{2}.$
See here for many more examples, and see this answer for general radical denesting algorithms.
Best Answer
Good question! You have discovered that it's not possible to define a square root function in the complex numbers that obeys the rule $\sqrt{ab}=\sqrt{a}\,\sqrt{b}$ (or the equivalent $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$, with $b\ne0$).
You get the same dilemma, in an easier way, by considering $$ i=\sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}=-i $$ Note that this is clearly wrong, which doesn't tell us that mathematics is contradictory, but that we have used an unproved (and unprovable) property, namely that we can define a square root function satisfying the rule above.
Note that the false argument produces both complex numbers whose square is $-1$, the same happens in your argument.
A suggestion: never use the symbol $\sqrt{-1}$, because it suggests the possibility to apply the wrong property. Neither use $\sqrt{z}$, for the same reason, unless $z$ is a real number with $z\ge0$.