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.
 A: The given equation is equivalent to $Z^3=-i$, indeed note simply that 
$$\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}\iff\bar Z-iZ^2=|Z|^2-1$$
then $\bar Z-iZ^2$ is real which requires


*

*-$iZ^2=Z\implies Z=0 \quad \lor \quad -iZ=1\implies Z=i$ ($Z=0$ is not a solution)


or


*

*$\bar Z-iZ^2=0\implies$ (from original equation) $|Z|=1$


thus the condition  $\bar Z-iZ^2$ is real requires $|Z|=1$ and the solution is given by
$$\bar Z-iZ^2=0\implies Z^3=-i \implies Z_1=i,\,Z_{2,3}=\frac{-i\pm\sqrt 3}{2}$$
A: The complex number $Z$ can be written as $$Z =  A + iB.$$
Then, your equation becomes:
$$\overline{Z} + 1 = iZ^2 + |Z|^2 \Rightarrow \\
A-iB + 1 = i(A^2-B^2+2iAB)+A^2+B^2 \Rightarrow \\
A-iB + 1 = iA^2 - iB^2 -2AB + A^2 + B^2 \Rightarrow \\
(A+1 +2AB-A^2-B^2) + i(-B-A^2+B^2) = 0.$$
This means that both the real part $(A+1 +2AB-A^2-B^2)$ and the imaginary part $(-B-A^2+B^2)$ must be $0$. Then, you must solve the following system:
$$\begin{cases}
A+1 +2AB-A^2-B^2 & = 0\\
-B-A^2+B^2 & = 0
\end{cases}. $$
This system has the following solutions:
$$A_1 = -\frac{\sqrt{3}}{2}, B_1 = -\frac{1}{2} \Rightarrow Z_1 = -\frac{\sqrt{3}}{2} -i\frac{1}{2},\\
A_2 = \frac{\sqrt{3}}{2}, B_2 = -\frac{1}{2} \Rightarrow Z_2 =  \frac{\sqrt{3}}{2}-i\frac{1}{2}, \\
A_3 = 0, B_3 = 1 \Rightarrow Z_3 = i.$$
A: \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}
$$z \in \left\{ i, 
                -\dfrac{\sqrt 3}{2} - i\dfrac 12,
                \dfrac{\sqrt 3}{2} - i\dfrac 12
        \right\} $$
