Question

Find the complex number for which |z+1| = z+2(1+i).

Answer 1

Let the required complex number be z=(x+iy). Then,
$|z+1|=z+2(1+i)$
$\Rightarrow |(x+iy)+1|=(x+iy)+2(1+i)$
$\Rightarrow \sqrt{(x+1{)}^2+y^2}=(x+2)+i(y+2)$
$\Rightarrow \sqrt{(x+1{)}^2+y^2}=(x+2)$ and y+2=0 [equating real parts and imaginary parts separately]
$\Rightarrow y=-2$ and $\sqrt{(x+1{)}^2+(-2{)}^2}=(x+2)$
$\Rightarrow y=-2$ and $\sqrt{x^2+2x+5}=(x+2)$
$\Rightarrow y=-2$ and $(x^2+2x+5)=(x+2{)}^2$
$\Rightarrow x^2+2x+5=x^2+4x+4$ and $y=-2$
$\Rightarrow 2x=1$ and $y=-2\Rightarrow x=\frac{1}{2}$ and $y=-2$.
Hence, the required complex number is $z=(\frac{1}{2}-2i)$.