What is geometric interpretation of $z \in \mathbb{C}$ such that $\left|\frac{z+1}{z-1}\right| < 1$?

  • 4
    $\begingroup$ $$\frac{|z+1|}{|z-1|}<1\implies |z+1|<|z-1|$$so these are points that are closer to $-1$ than $1$. $\endgroup$ – Dave Oct 22 '17 at 22:06
  • $\begingroup$ It's a complicated way of saying $Re(z)<0$. $\endgroup$ – J. Salieri Oct 22 '17 at 22:53

Write $z = x+iy$ and expand out your inequality in terms of $x$ and $y$. What do you get?


Correct me if wrong:

Consider $(+1,0)$ and $(-1,0)$ in the complex plane.

Any $z\in \mathbb{C},$ I.e. $z =x +iy,$ or $z=(x,y).$

$|z+1|$ is the distance form

$z(x,y)$ to $(-1,0)$.

$|z-1|$ is the distance from

$z(x,y)$ to $(1,0).$

The locus of points such that

$\dfrac{|z+1|}{|z-1|} = r$ , real and positive, are the

Appolonius' Circles.


$ r \lt 1$ are the blue circles to the left of

the perpendicular bisector of

$ (-1,0)$ and $(1,0).$


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