# Geometrical interpretation ($\left| \frac{z-1}{z-i}\right| > 1$, $\arg z < \pi$)

I'd like to ask you about the example below (I have to draw a geometrical interpretation in an argand diagram).

$\left| \frac{z-1}{z-i}\right| > 1$, $\arg z < \pi$

What I did so far:

$|z-1| > |z-i|$

$|x + yi -1| > |x + yi -i|$

$\sqrt{(x-1)^2 + y^2} > \sqrt{x^2 + (y-1)^2}$

$(x-1)^2 + y^2 > x^2 + (y-1)^2$

$x^2 - 2x + 1 +y^2 > x^2 + y^2 -2y + 1$

$- 2x + 1>-2y + 1$

$-x +y > 0$

$y > x$

And now... The question is:

How to draw it? Is it the surface over $y=x$ function? What about the $\arg z < \pi$? Does it matter in this case? Or: • What's the range for $\operatorname{arg}z$? Is it $(-\pi, \pi]$, $[0, 2\pi)$ or something else? – Henricus V. Oct 18 '15 at 0:20
• @Henry W, RegionPlot[y > x && Arg[x + I y] <= Pi, {x, -2, 2}, {y, -2, 2}] that was the original query. The only condition was $\arg z < \pi$. – belford Oct 18 '15 at 0:22
• @HenryW, what do you think about the second one? I've redefined the $\arg z$. myArg[z_] := Mod[Arg@z, 2 Pi] RegionPlot[y > x && myArg[x + I y] <= Pi, {x, -2, 2}, {y, -2, 2}] – belford Oct 18 '15 at 0:44
• I don't know how the $\operatorname{mod}$ function in mathematica would behave for negative values, but I can tell you that $\operatorname{arg}$ is always between $-\pi$ and $\pi$. – Henricus V. Oct 18 '15 at 0:48
• Is $\arg z < \pi$ condition unnecessary? It was originally in the exercise. I thought that $\arg z \in < 0, 2\pi)$ and then the second image should be fine? – belford Oct 18 '15 at 0:53

There was no need to do so much computation. $|z-1| > |z-i|$ is equivalent to $z$ being further from $1$ than $i$, and hence it is clearly the open region bounded by the perpendicular bisector of the points $(1,i)$. The other condition says that it is in the open half plane above the real axis. Combine the two to get the desired region. (So you should be using dotted lines rather than solid lines by convention.)