1. ## mobius transform question

$\displaystyle D^+=\left \{z:|z|<1,Im \left \{ z \right \}>0 \right \}$
so it represents the northen hemisphere of a circle with radius 1.
$\displaystyle f(z)=\frac{2z-i}{2-iz}$
i need to find what is the picture of $\displaystyle f(d^+)=?$

i tried to solve it like this:
my area is bounded by a line and a curve.
i want to see what each one transforms to
curve points:
f(1)=1
f(-1)=-1
f(i)=i/3
so it will look like this
http://i46.tinypic.com/3129rg4.gif

line points:
f(1)=1
f(-1)=-1
f(0)=-i/2
so it looks like this
http://i49.tinypic.com/dnnh2u.jpg

and when i try and see where the inside goes :
f(i/2)=0
so the answer represents their intersection area
(i mistakenly marked only a part of the interssection marked it should be the whole intersection)
http://i49.tinypic.com/14bi2qr.gif

but my prof thinks otherwise
http://i48.tinypic.com/2jff52r.jpg

who is worng here? why?
so its in their intersections

2. Originally Posted by transgalactic
$\displaystyle D^+=\left \{z:|z|<1,Im \left \{ z \right \}>0 \right \}$
so it represents the northen hemisphere of a circle with radius 1.
$\displaystyle f(z)=\frac{2z-i}{2-iz}$
i need to find what is the picture of $\displaystyle f(d^+)=?$

i tried to solve it like this:
my area is bounded by a line and a curve.
i want to see what each one transforms to
curve points:
f(1)=1
f(-1)=-1
f(i)=i/3
so it will look like this
http://i46.tinypic.com/3129rg4.gif

line points:
f(1)=1
f(-1)=-1
f(0)=-i/2
so it looks like this
http://i49.tinypic.com/dnnh2u.jpg

and when i try and see where the inside goes :
f(i/2)=0
so the answer represents their intersection area
(i mistakenly marked only a part of the interssection marked it should be the whole intersection)
http://i49.tinypic.com/14bi2qr.gif

but my prof thinks otherwise
http://i48.tinypic.com/2jff52r.jpg

who is worng here? why?
so its in their intersections

Put $\displaystyle z=x+iy\,,\,\,so\,\,\,z\in\ D^+\Longrightarrow x^2+y^2<1\,,\,\,y>0$ , and from here:

$\displaystyle \frac{2z-i}{2-iz}=\frac{5x}{x^2+(y+2)^2}+\frac{2\left[x^2+(y+2)\left(y-\frac{1}{2}\right)\right]}{x^2+(y+2)^2}\,i$

It's easy to see that the imaginary part above can be negative, for example take $\displaystyle z=0.1+0.1i\in D^+$, so the image cannot be what your prof says since that image has all its imaginary part positive...

Tonio

3. thanks