Is $|z-i| = |z+i|$? I computed a Mobius transformation $-\frac{z-i}{z+i}$ that maps the upper half plane to a disk, with i mapping to the center of the disk, $w = 0$.
How do I know that the disk is a unit disk and not anything bigger or smaller? 
I tried using the definition of modulus: $w = f(z) = |\frac{z-i}{z+i}| = \sqrt{\frac{x^2+(y-1)^2} { x^2+(y+1)^2}}$, but does this equal $1$?
Or, better still, is $|z-1| = |z+1|$ and $|z-2| = |z+2|$, etc.  Is there such a relation?
Thanks,
 A: In fact, $$\vert z+i\vert=\vert z-i\vert$$ is satisfied by no $z\in\mathbb C\setminus\mathbb R.$ Square each side to get $$(z+i)(\bar z-i)=(\bar z+i)(z-i)$$ which expands to $$z\bar z+i(\bar z-z)+1=z\bar z+i(z-\bar z)+1$$ or $$\bar z=z\implies z\in\mathbb R.$$ General relations like $$\vert z+a\vert=\vert z-a\vert$$ if $a\in\mathbb C$ can be solved similarly: we get $$(z+a)(\bar z-\bar a)=(z-a)(\bar z+\bar a)$$ or $$a\bar z-\bar az=0$$ after some manipulation. What number must be real for this to hold?
A: Hint: to compare $|z-a|$ and $|z+a|$, what happens when $z=a$ (or $z=-a$)? Can there be any meaningful relationship between them?
A: The locus of points that are equidistant from two points is the perpendicular bisector of the segment connecting the two points. The locus of points so that $|z+i|=|z-i|$ is the set of points on the perpendicular bisector of the segment connecting $-i$ and $i$; that is, the real axis. Thus, the locus of points so that $\left|\frac{z+i}{z-i}\right|=1$ is the real axis. In other words, $\frac{z+i}{z-i}$ is on the unit circle if and only if $z$ is real.
