Find the roots of $\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}$ I attempted the problem in two ways     
Method 1
Resolved $\mathrm{ Z = X + iY}$ this led to a very big cubic equation and handling that became too difficult for me .   
Method 2
I believe factorizing the equation is a far better method, I don't know how to go about it .

EDIT
The answer given is $\mathrm{Z = \frac{-i}{2} , i}$  
As pointed out by @gimsui $\mathrm{\frac{-i}{2} }$ is not a solution most likely a printing error.

\begin{align}
   z &= x + iy \\
\hline
   \overline z &= x - iy \\
   z^2 &= (x^2-y^2)+2ixy \\
   |z|^2 &= x^2+y^2
\end{align}
\begin{align}
   \overline z + 1 &= iz^2 + |z|^2 \\
   (x-iy) + 1 &= i(x^2-y^2 + 2ixy)+(x^2+y^2) \\
   (x+1) - iy &= (x^2-2xy+y^2) + i(x^2-y^2) \\
   x+1-x^2+2xy-y^2 &= i(x^2-y^2+y) \\
   \hline
   x^2-2xy+y^2 - x - 1 &= 0 \\
   x^2 &= y^2-y \\
   \hline
   2y^2-2xy-x-y-1 &= 0 \\
   2y^2 -(2x+1)y-(x+1)&=0 \\
   y &= \dfrac{(2x+1)\pm \sqrt{(2x+1)^2+8(x+1)}}{4} \\
   y &= \dfrac{(2x+1)\pm \sqrt{4x^2+12x+9}}{4} \\
   y &= \dfrac{(2x+1)\pm |2x+3|}{4} \\
   y &= \dfrac{(2x+1)\pm (2x+3)}{4} \\
   y &\in \left\{ x+1, -\frac 12 \right\}
\end{align}
If $y=x+1$ and $x^2=y^2-y$, then
\begin{align}
   x^2 &= (x+1)^2 - (x+1) \\
   x^2 &= x^2 + x \\
   (x,y) &= (0,1)
\end{align}
If $y=-\dfrac 12$ and $x^2=y^2-y$, then
\begin{align}
   x^2 &= \dfrac 14 + \dfrac 12 \\
   (x,y) &= \left( \pm \dfrac{\sqrt 3}{2}, -\dfrac 12 \right)
\end{align}
$$z \in \left\{ i, 
                -\dfrac{\sqrt 3}{2} - i\dfrac 12,
                \dfrac{\sqrt 3}{2} - i\dfrac 12
        \right\} $$
