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.

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:
$$A_1 = -\frac{\sqrt{3}}{2}, B_1 = -\frac{1}{2} \Rightarrow Z_1 = -\frac{\sqrt{3}}{2} -i\frac{1}{2},\\
A_2 = \frac{\sqrt{3}}{2}, B_2 = -\frac{1}{2} \Rightarrow Z_2 =  \frac{\sqrt{3}}{2}-i\frac{1}{2}, \\
A_3 = 0, B_3 = 1 \Rightarrow Z_3 = i.$$
