Isomorphisms of Complex Regions: Finding Corresponding Functions

[WannaBe22]

Homework Statement

1. Find and isomorphism of the region |arg z|< \frac{\pi}{8} on the region:
|z|<1,z Not In [0,1).

2. Find and isomorphism of the region D= (z=i+re^{it}, \frac{5 \pi}{4} < t<\frac{7 \pi}{4}, r \geq 0 ) on the region R=(z: 0<Rez< \pi , Imz>0).

Homework Equations

Mobiuos Transformations...Elementry Functions

The Attempt at a Solution

I have no idea about 2...
About 1- I thought about first taking the function z \to z^8 which will copy the region to C-(- \infty, 0] ... I was thinking about using LOG or something from that point, but I have no idea how to continue this composition of functions

Hope you'll be able to help me

Thanks

 

[tmccullough]

I'll give you a hint to point you in the right
direction for both of these. The second one is
just one additional step from the first.

A basic beginning for both of these problems is
to begin by mapping the region in question to
the region R = Re z > 0, I am z > 0.

For 1, the function is f(z) = z^2*exp(i*pi/4).

Now, consider applying g(z) = 1/(z+1), which maps
the I am z = 0,Re z > 0 boundary of R to the line
segment (0,1) and the Re z = 0, I am z > 0
boundary of R to the upper half of the circle
|z-1/2| = 1/2...do you see that? So, g maps R
onto the upper half of the disk |z-1/2| < 1/2.

Then, note that h(z) = (2z-1)^2 maps this upper
half disk onto |z| < 1, z not in [0,1).

Then, the whole function is the composition h(g(f(z))).

For 2, recall that exp(z) is an isomorphism from the
horizontal strip 0 < I am z < 2pi, Re z < 0 onto
|z| < 1, z not in [0,1).

Then, the so called principle branch of log(z) is an
isomorphism in the opposite direction. Now, proceed
as for 1 to find an isomorphism from the region onto
|z| < 1, z not in [0,1), then compose with log(z).
Finally, you have to rotate and compress (multiply
by 1/2*exp(-i*pi/2)).

Hopefully, that helps.

 

[WannaBe22]

Wow! Thanks a lot for the detailed guidance! It was awsome...
Unfortunately, there is one thing I can't understand:

Now, consider applying g(z) = 1/(z+1), which maps
the I am z = 0,Re z > 0 boundary of R to the line
segment (0,1) and the Re z = 0, I am z > 0
boundary of R to the upper half of the circle
|z-1/2| = 1/2...do you see that? So, g maps R
onto the upper half of the disk |z-1/2| < 1/2.

1. As far as I know, when we take the map \frac{1}{z} , it mappes I am z = 0,Re z > 0 to (0,1)... When we take the map \frac{1}{z+1} , it mappes this region to (1,2)...Am I right?
2. I can't figure out why this map copies I am z > 0,Re z = 0 to the upper half of the circle
|z-1/2| = 1/2 ... This is the part I couldn't understand in your guidance...
I thought the map \frac{1}{z} takes the first quadrant to :
(z: |z|&lt;1 , Imz&gt;0,Rez&gt;0) ... So the map \frac{1}{z+1} takes this region to the circle:(z: |z-1|&lt;1 , Imz&gt;0,Rez&gt;0) ... I really can't understand where is my mistake in understanding these mappings... Hope you'll be able to help


Thanks a lot again

 

[tmccullough]

First, sorry that I crapped out of texifying...my first time posting here, and I didn't know how to do it.

Note that if z\in (0,1) then \frac{1}{z} \in (1,\infty), and z\in (1,\infty) then \frac{1}{z} \in (0,1), so you are not quite right.

This is the same reason that \frac{1}{z+1} maps (0,\infty) \longrightarrow (0,1)

Second, remember that Mobius transformations map lines to circles. The vertical line Re z = 0 gets mapped to some kind of circle by \frac{1}{z+1}. Note 0\longrightarrow 1 and \infty\longrightarrow 0.

Hopefully, it is clear why Re z = 0, I am z &gt; 0 maps to half of the circle. I actually mispoke, it's the bottom half - figure out why and go from there. You're welcome - it was a welcome diversion, I'm deployed in Iraq right now.

 

[WannaBe22]

I think it's completely understandble now...Thanks again!