<s>[INST] Solving $\sqrt{i}$ The results I get not the same as in the book, basically I need to get only this and the $x$'s of course: 
$$y = \frac{\sqrt{2}}{2}$$
So this is what I did:
$$\sqrt{i} = x+yi$$
$$i = x^2 +2xyi -y^2$$
$$\begin{cases} x^2-y^2 = 0 \\ 
2xy = 1 \end{cases}$$
$$xy=0.5$$
$$x=\frac{0.5}{y}$$
$$(1)\space\space\space\space(\frac{0.5}{y})^2-y^2=0$$
$$0.25-y^4=0$$
$$y^4=0.25$$
$$y=\pm \sqrt[4]{0.25}$$
$$y=\pm \frac{\sqrt{2}}{2}$$
What is WRONG?
 [/INST] You have $y = \pm \dfrac{\sqrt 2} 2$.
If $y= \dfrac{\sqrt 2} 2$ then $x=\dfrac{0.5} y = \dfrac{0.5}{\sqrt 2/2}$, and this simplifies to $x=\dfrac{\sqrt 2} 2$.
So $\dfrac{\sqrt 2} 2 + i \dfrac{\sqrt 2} 2$ is a complex number whose square is $i$.
If you multiply that by $-1$ you get the other complex number whose square is $i$.  That is the one that comes from $y=\dfrac{-\sqrt 2} 2\vphantom{\dfrac{\displaystyle\int}\int}$.
One can show that there cannot be more than two solutions: Recall from algebra that if $a$ is a solution to $x^2 -i=0$ then $x-a$ is a factor of $x^2-i$, so you get
$$
x^2 - i = (x-a)(\cdots\cdots).
$$
The other factor must be a first-degree polynomial, so it can't have more than one root.
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