# Mapping complex numbers

• Mar 15th 2011, 07:42 PM
Glitch
Mapping complex numbers
The question:

For the mapping f(z) = \frac{z − i}{z + i}, ﬁnd the image of {z ∈ C : Im(z) ≥ 0}

I'm confused as to how to attempt this. I tried setting f(z) = w, the rearranged to make z the subject. I then set w = a + ib and found the imaginary part. But it was just a big mess, and I'm fairly sure I have no idea what I'm doing. My text is quite brief on this topic.

Any assistance would be great.
• Mar 15th 2011, 08:21 PM
xxp9
If z is in the upper half plane, z is closer to i than (-i). So |z-i|<=|z+i|, so |(z-i)/(z+i)| <= 1, so f(z) is in the unit disk. And the image of the upper half plane is the unit disk.
• Mar 15th 2011, 09:16 PM
FernandoRevilla
In general, all the of Möbius maps $f$ transforming $\textrm{Im}z>0$ in $|z|<1$ , with $f(z_0)=0$ are:

$f(z)=e^{i\alpha}\dfrac{z-z_0}{z-\bar{z_0}}\quad(\alpha\in\mathbb{R})$

In our case, $z_0=i,\;\alpha=0$ .
• Mar 16th 2011, 12:53 AM
Glitch
Thanks guys, took me a while but I got it. Man this stuff is difficult. >_<
• Mar 16th 2011, 01:07 AM
Glitch
Oh, and the solution says "the closed disc |w| <= 1 with w = 1 removed". Why is w = 1 removed? Thanks.
• Mar 16th 2011, 01:34 AM
FernandoRevilla
Because $f(z)=1\Leftrightarrow z=\infty$. For that reason I chose $\textrm{Im}z>0$ instead of $\textrm{Im}z\geq 0$ .
• Mar 16th 2011, 01:41 AM
Opalg
Quote:

Originally Posted by Glitch
Oh, and the solution says "the closed disc |w| <= 1 with w = 1 removed". Why is w = 1 removed? Thanks.

That's a very nice little trick in xxp9's comment. It gets the solution very slickly, but it misses the fact that w=1 is omitted. To see that, it's better to use a less imaginative method.

If $w = \frac{z - i}{z + i}$ then $z = \frac{i(1+w)}{1-w} = \frac{i(1+w)(1-\overline{w})}{(1-w)(1-\overline{w})} = \frac{i(1-|w|^2 +w-\overline{w})}{|1-w|^2}$ (where the bar denotes the complex conjugate). Since $w-\overline{w}$ is purely imaginary, the imaginary part of z is $\frac{1-|w|^2}{|1-w|^2}$. The numerator tells you that if $\text{Im}(z)\geqslant0$ then $|w|\leqslant1$. But also, the denominator has to be nonzero, which tells you that $w\ne1$.
• Mar 16th 2011, 02:49 AM
Glitch
Thanks Opalg, that looks similar to my first attempt.
• Mar 16th 2011, 05:04 AM
xxp9
Right I didn't show that the map is surjective or not. Actually there is still a nice way to visualize. The inverse of the stereographic projection sends the upper half plane to the half sphere.
Now check f(z)=(z-i)/(z+i)=[i(z-i)/(z+i)] * [(-i)] = h( g(z) )
On the sphere, g(z)=i(z-i)/(z+i) sends i to 0( south pole), -i to infinity( north pole), 0( south pole) to -i, infinity( north pole) to i. g is the rotation by 90 along the real axis, with -1 and 1 fixed.
And h(z) = -iz rotates the sphere along the z-axis, keeping the south and north poles fixed, sending i to 1 and -i to -1.
In this process, the upper half plane is
1) sent to the half sphere first,
2) then rotated to south half sphere by g,
3) then this south half sphere is rotated along the z-axis by h, keeping in south sphere.
4) Then the stereographic projection sends the south sphere to the unit disk.
To see that 1 is omitted, note that the first step 1) is not surjective. The north pole is never covered.