Question

The points representing the complex number z for which |z + 1| < |z – 1| lie on the left side of ____________.

Answer 1

Given |z + 1| < |z – 1|

Let z = x + iy where x, y ∈R

$$\begin{gathered} \mathrm{i}.\mathrm{e}\left|x+iy+1\right|<\left|x+iy-1\right| \\ \Rightarrow \left|\left(x+1\right)+iy\right|<\left|\left(x-1\right)+iy\right| \end{gathered}$$

Squaring both sides, we get

$$\begin{gathered} {\left|\left(x+1\right)+iy\right|}^2<{\left|\left(x-1\right)+iy\right|}^2 \\ \mathrm{i}.\mathrm{e}{\left(x+1\right)}^2+y^2<{\left(x-1\right)}^2+y^2 \\ \mathrm{i}.\mathrm{e}{x}^2+1+2x+y^2<x^2+1-2x+y^2 \\ \mathrm{i}.\mathrm{e}2x<0 \\ \mathrm{i}.\mathrm{e}x<0 \end{gathered}$$

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