If

Solution:

Let $z=x+i y$.

Then,

$z+1=(x+1)+i y$

$\Rightarrow|z+1|=\sqrt{(x+1)^{2}+y^{2}}$

$\therefore|z+1|=z+2(1+i)$

$\Rightarrow \sqrt{x^{2}+2 x+1+y^{2}}=(x+i y)+2+2 i$

$\Rightarrow \sqrt{x^{2}+2 x+1+y^{2}}=(x+2)+i(y+2)$

$\Rightarrow \sqrt{x^{2}+2 x+1+y^{2}}=(x+2)$ and $y+2=0$

$\Rightarrow x^{2}+2 x+1+y^{2}=(x+2)^{2}$ and $y=-2$

$\Rightarrow x^{2}+2 x+1+y^{2}=x^{2}+4 x+4$ and $y=-2$

$\Rightarrow y^{2}=2 x+3$ and $y=-2$

$\Rightarrow 4=2 x+3$ and $y=-2$

$\Rightarrow 2 x=1$ and $y=-2$

$\Rightarrow x=\frac{1}{2}$ and $y=-2$

$\therefore z=x+i y=\frac{1}{2}-2 i$

Thus, $z=\frac{1}{2}-2 i$