Question

If $|z+1|=z+2(1+i)$
find z.

Answer 1

Let $z=x+iy$
$\because |z+1|=z+2(1+i)$
$\Rightarrow |x+iy+1|=x+iy+2(1+i)$
$\Rightarrow \sqrt{(x+1{)}^2+y^2}=(x+2)+i(y+2)$
Equate the real and imaginary parts,
$\Rightarrow \sqrt{(x+1{)}^2+y^2}=(x+2)\dots \left(1\right)$
and$y=-2\cdots \left(2\right)$
Put value of 𝑦 in equation (1)
$\therefore \sqrt{(x+1{)}^2}+(-2{)}^2)=(x+2)$
$\Rightarrow (x+1{)}^2+4=(x+2{)}^2$
{squaring both of side}
$\Rightarrow x^2+2x+1+4=x^2+4x+4$
$\Rightarrow x=\frac{1}{2}$
Therefore, $z=\frac{1}{2}-2i$