The points representing the complex number z for which

Given |z + 1| < |z – 1|

Let z = x + iy where x, y ∈R

i. e $|x+i y+1|<|x+i y-1|$

$\Rightarrow|(x+1)+i y|<|(x-1)+i y|$

Squaring both sides, we get

$|(x+1)+i y|^{2}<|(x-1)+i y|^{2}$

i.e $(x+1)^{2}+y^{2}<(x-1)^{2}+y^{2}$

i. e $x^{2}+1+2 x+y^{2}

i. e $2 x<0$

i. e $x<0$

Hence, $|z+1|<|z-1|$ lies on left side of $y$-axis