$$
\begin{aligned}
&\text { We have }|z+1|=z+2(1+i)\\
&\text { Let } z=x+i y\\
&\text { Then, }|x+i y+1|=x+i y+2(1+i)\\
&\Rightarrow \quad|x+1+i y|=(x+2)+i(y+2)\\
&\Rightarrow \sqrt{(x+1)^2+y^2}=(x+2)+i(y+2)\\
&\left[\because|z|=\sqrt{x^2+y^2}\right]\\
&\text { On squaring both sides, we get }\\
&(x+1)^2+y^2=(x+2)^2+i^2(y+2)^2+2 i(x+2)(y+2)\\
&\Rightarrow x^2+2 x+1+y^2=x^2+4 x+4-y^2-4 y\\
&-4+2 i(x+2)(y+2)\\
&\Rightarrow x^2+y^2+2 x+1=x^2-y^2+4 x-4 y\\
&+2 i(x+2)(y+2)
\end{aligned}
$$
On comparing real and imaginary parts, we get
$x^2+y^2+2 x+1=x^2-y^2+4 x-4 y$
$\Rightarrow \quad 2 y^2-2 x+4 y+1=0$
and $\quad 2(x+2)(y+2)=0$
$\Rightarrow \quad x+2=0$ or $y+2=0$
$x=-2$ or $y=-2$
For $x=-2, \quad 2 y^2+4+4 y+1=0$
[using Eq. (ii)]
$\Rightarrow 2 y^2+4 y+5=0$
So $D=b^2-4 a c=16-4 \times 2 \times 5=-24 < 0$
$\Rightarrow \quad 2 y^2+4 y+5$ has no real roots.
For $y=-2, \quad 2(-2)^2-2 x+4(-2)+1=0$
[using Eq. (ii)]
$\Rightarrow \quad 8-2 x-8+1=0$
$\Rightarrow \quad x=1 / 2$
$\therefore \quad z=x+i y=\frac{1}{a}-2 i$