Find the complex number z for which

Solution:

Given: $|z|=z+1+2 i$

Consider,

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

Squaring both the sides, we get

$|z|^{2}=[(z+1)+(2 i)]^{2}$

$\Rightarrow|z|^{2}=|z+1|^{2}+4 i^{2}+2(2 i)(z+1)$

$\Rightarrow|z|^{2}=|z|^{2}+1+2 z+4(-1)+4 i(z+1)$

$\Rightarrow 0=1+2 z-4+4 i(z+1)$

$\Rightarrow 2 z-3+4 i(z+1)=0$

Let $z=x+i y$

$\Rightarrow 2(x+i y)-3+4 i(x+i y+1)=0$

$\Rightarrow 2 x+2 i y-3+4 i x+4 i^{2} y+4 i=0$

$\Rightarrow 2 x+2 i y-3+4 i x+4(-1) y+4 i=0$

$\Rightarrow 2 x-3-4 y+i(4 x+2 y+4)=0$

Comparing the real part, we get

$2 x-3-4 y=0$

$\Rightarrow 2 x-4 y=3 \ldots(i)$

Comparing the imaginary part, we get

$4 x+2 y+4=0$

$\Rightarrow 2 x+y+2=0$

$\Rightarrow 2 x+y=-2 \ldots$ (ii)

Subtracting eq. (ii) from (i), we get

$2 x-4 y-(2 x+y)=3-(-2)$

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

$\Rightarrow-5 y=5$

$\Rightarrow y=-1$

Putting the value of $y=-1$ in eq. (i), we get

$2 x-4(-1)=3$

$\Rightarrow 2 x+4=3$

$\Rightarrow 2 x=3-4$

$\Rightarrow 2 x=-1$

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

Hence, the value of $z=x+i y$

$=-\frac{1}{2}+i(-1)$

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