Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
     \frac{x^2-y^2}{2} = 0  \\
     xy = 1\\   \end{cases}   \\
=\begin{cases}
     x^2-y^2 = 0  \\
     x = \frac{1}{y}\\   \end{cases}   \\
=\begin{cases}
     \frac{1}{y}-y^2 = 0  \\
     x = \frac{1}{y}\\   \end{cases}   \\=
\begin{cases}
     \frac{1-y^3}{y} = 0  \\
     -\\   \end{cases}   
\\=
\begin{cases}
     y^3 = 1  \\
     -\\   \end{cases}  
\\=
\begin{cases}
     y = 1  \\
     x =1\\   \end{cases}    $$
And so $\sqrt{2i} = 1+i$, but my book states the solution is $\sqrt{2i} = 1+i$ and $\sqrt{2i} = -1-i$.
What did I forget?
 A: $2i=1+2i+i^2=(1+i)^2$. The roots are $\pm(1+i)$.
A: In complex numbers, square roots (and cube roots, and fourth roots, etc) are no longer uniquely defined; a square root of a complex number is any value that, when squared, yields the number.
So if one value of $\sqrt{z}$ is $w$, then another value is $-w$, since $w^2 = (-w)^2 = z$.
A: Write $2i=2e^{i \pi/2+2k\pi}$. Then square root to get: $\sqrt{2} e^{i\pi/4+k\pi}$. So your roots are $\sqrt{2}e^{i\pi/4}$ and $\sqrt{2}e^{3i\pi/4}$. Which are $\pm(1+i)$.
A: I'll stick with your solving algorithm. You obtained (apart from a mistake):
$$\frac{1}{y^2}-y^2 = 0, \implies \frac{1}{y^2} = y^2$$
You can check that both $1$ and $-1$ satisfy this equation! So your chain of systems splits in two: knowing $x = \frac{1}{y}$,
$$y=1 \implies x = 1 \qquad y=-1 \implies x= -1 $$
So you get (as one should expect according to the Fundamental Theorem of Algebra) the two solutions
$$\sqrt{2i} = \pm(1+i) $$
Of course, you could have avoided this lengthy procedure in a few ways, for example by taking advantage of the polar form of $2i$. Take a peek at the other answers!
