Complex Numbers: How do I prove that the statements $|z+1|>|z−1|$ and Re(z)>0 are equivalent?

Question

I know that $|z + 1|$ represents the distance between $z$ and $-1$ and $|z - 1|$ represents the distance between $z$ and $1$. From here, I can derive that there will be a line $x$ that will be the perpendicular bisector of the line segments joining the points $(-1,0)$ and $(1,0)$. How do I continue from here?

Answer 1

\begin{align} |z+1|\gt|z-1| &\iff|z+1|^2\gt|z-1|^2\\ &\iff(\Re{(z)}+1)^2+\Im^2{(z)}\gt(\Re{(z)}-1)^2+\Im^2{(z)}\\ &\iff\Re^2{(z)}+2\Re{(z)}+1\gt\Re^2{(z)}-2\Re{(z)}+1\\ &\iff4\Re{(z)}\gt0\\ &\iff\Re{(z)}\gt0\\ \end{align}

Answer 2

The line $\Re(z)=0$ is the locus of points equidistant from $\pm1$, i.e. $|z+1|=|z-1|$. In particular, $|z+1|>|z-1|$ means $z$ is closer to $1$, by being on the same side of that line as $1$ is, which is equivalent to $\Re(z)>0$.

Answer 3

Let $z=x+iy$ where $x,y$ are real

We need $$(x+1)^2+y^2>(x-1)^2+y^2\iff4x>0$$

Answer 4

Just keep it simple.

$z = x+yi$

Given $|z+1|>|z-1|$,this is equivalent to $|z+1|^2>|z-1|^2$ (squaring is permitted as both sides are non-negative).

$|z+1|^2 = (x+1)^2 + y^2$

Similarly, $|z-1|^2 = (x-1)^2 + y^2$

Hence the condition reduces to $(x+1)^2 > (x-1)^2$

Don't expand, use the difference of squares identity:

$(x+1 - (x-1))(x+1 + x-1)> 0$

$2(2x)>0$

$x>0$

$\therefore \Re (z) >0$ (QED)

Note that all steps are reversible so the implication holds both ways.