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?

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 $\pm(1+i)$.
