I need to solve:

$$z^2+(1+i)z+i=0$$

first of all I used $$z = (a+ib)$$ and I get:

$$(a+ib)^2+(1+i)(a+ib)+i=0$$
$$a^2-b^2+2aib+a+ib+ia-b+i=0$$

then I have ordered, on the left the number without the $i$ and on the right all the number with the $i$:

$$\underbrace{a^2-b^2+a-b}\_\text{real} + \underbrace{2aib+ib+ia+i}\_\text{imaginary} = 0$$

So I have set this:

$$\begin{cases} a^2-b^2+a-b=0 \\ 2aib+ib+ia+i \end{cases}$$

$$\begin{cases} a(a+1)-b(b-1) = 0 \\ 2aib+ib+ia+i \end{cases}$$

How can I procede? I can say that the first equation is true when $a=b$