# Homework Help: Find analitical function f that maps conformal

1. Mar 28, 2014

### skrat

1. The problem statement, all variables and given/known data
Find analitical function $f$ that maps half-plane $Re(z)<-3$ conformal on:
a) half-plane $Re(z)>0$
b) half-plane $Im(z)>0$
e) open unit circle

2. Relevant equations

3. The attempt at a solution

I am somehow convinced that my solutions are wrong, please correct me if necessary and help me with e).

a): $f(z)=|z|-3$

b): $f(z)=i(|z|-3)$

c): $f(z)=|z|-3+i(|z|-3)$

d): $f(z)=|z|+3+i(|z|-3)$

e): ....?

2. Mar 28, 2014

### LCKurtz

They are all wrong because |z| is not analytic. I would suggest thinking about rotations and translations to get started.

3. Mar 28, 2014

### skrat

Does the answer to part a) sound something like $f(z)=-z-3$?

If $z=-3$ than $f(z)=0$ which is fine.
If $z=-4$ than $f(z)=1$ which is also fine.
If $z=-4+2i$ than $f(z)=1-2i$ which is still in $Re(z)>0$ half-plane.

So $f(z)$ maps $Re(z)<-3$ to $Re(z)>0$ and rotates it around $Im$ axis.

Also $f(z)^{'}=-1 \neq 0$ so $f(z)$ is conformal.

4. Mar 28, 2014

### LCKurtz

That does map it to $Re(z)>0$ conformally. But I wouldn't call any part of it a rotation about the imaginary axis. More like a reflection ($-z$) and translation $-3$. Still, it is correct for a.

 On second thought, $-z$ can be viewed as a rotation or a reflection of that region.

Last edited: Mar 28, 2014
5. Mar 28, 2014

### skrat

Ok.

than b) $f(z)=-i(z-3)$ ; that's if it is ok that $f(z)^{'}=-i$ ?

c)??

Is there a way to actually calculate these things or is this something one has to simply "see"...?

6. Mar 28, 2014

### LCKurtz

That is close but isn't quite correct. For example, no $z$ in the region maps to $i$. You can see that because if $i = -i(z-3)$ you get $z=2$ for the point that maps to $i$. But that isn't in your original region. Think about the translation again.

For the first parts of your problem you should be thinking about translations and rotations. Do you know how to rotate the plane about the origin? You can always translate to the origin, rotate it, and translate it again if necessary.

7. Mar 28, 2014

### skrat

How on earth did you see that one point that isn't mapped?

I can rotate it with multiplying it with $e^{i\varphi }$. Give me some time, I hope I will come back with something useful if not, I will just have to hit myself hard.

8. Mar 28, 2014

### LCKurtz

It was easy. I just looked geometrically at what you did. $z-3$ translates the region 3 units to the left and multiplication by -i rotates it. Not what you wanted to do. If you understand why it's wrong you will see how to fix it.

9. Mar 28, 2014

### skrat

b) from $Re(z)<-3$ to $Im(z)>0$

$f(z)=e^{-i\pi /2}z +3$

Or is the translation completely unnecessary here?

10. Mar 28, 2014

### LCKurtz

Nope. That rotates it about the origin then moves it to the right 3.

11. Mar 28, 2014

### skrat

than $f(z)=ze^{-i\pi /2 }$.

12. Mar 28, 2014

### skrat

Aha no. I take that last one back. :D

13. Mar 28, 2014

### LCKurtz

Tell me in words, not equations, what you have to do geometrically to get the region where you want it.

14. Mar 28, 2014

### skrat

I want all the vertical lines: $a+it$ for $t\in[-\infty , \infty]$ and $a<-3$ to be mapped into lines $t+ia$ where $t\in[-\infty , \infty]$ and $a>0$.
In words: I want all the vertical lines to be mapped into horizontal lines above 0.

Therefore it would be nice to rotate them for $-\pi /2$ so they become horizontal and than move them vertically if necessary.

So $f(z)=ze^{-i\pi /2}-3=|z|e^{i(\varphi -\pi/2)}-3$.

For given $\varphi$ and all $|z|>3$ the lines will be rotated for $-\pi /2$ backwards, therefore all the lines will be on the upper half of Imaginary axis. However I have to move them for 3 units down, that is why I used -3 in equation.

15. Mar 28, 2014

### LCKurtz

Yes. Or you could just say rotate the region about the origin.

Arghhh. No $|z|$ allowed. See post #2. But that $-3$ moves the region 3 units to the left. Not what you want. You want to move it 3 units down after you rotate it. It's an easy fix though.

Alternatively, you could first move the region 3 units to the right before rotating. You should correct the above then try that too. You will get the same answer and it's a learning experience.

I have to go now. Back in a few hours. Good luck.

16. Mar 28, 2014

### skrat

Now I can see why people have problems with complex analysis. Just after you think you have it all figured out a simple problem like this comes and hits you hard.

$f(z)=ze^{-i\pi /2}-3i$

I completely forgot that I am moving the lines on Imaginary axis, therefore an $i$ was missing. However, this should be it. Rotation for $-\pi /2$ and translation for 3 units lower on Imaginary axis.

I need some rest now. This was too exhausting. Thanks for all the help!

17. Mar 28, 2014

### skrat

From $Re(z)<-3$ to

c) Firstly I want to move all the lines to origin and than I change my view to half circles. I want them to be a quarter of a circle about the origin and i can squeeze them using $z^{1/2}$.
However, line at $\pi /2$ will now go to $\pi /4$, meaning I have to rotate it for extra $-\pi /4$.

Therefore $f(z)=(z+3)^{1/2}e^{-i\pi /4}$

and accordingly $f(z)=(z+3)^{1/2}e^{i\pi /4}$ for d) part.

for e) unit circle, I imagine I have to map all the vertical lines into circles inside an open unit circle.

18. Mar 28, 2014

### LCKurtz

Yes. Notice that $e^{-i\pi /2}=-i$ so you could write it as $-iz-3i$ or $-i(z+3)$. Do you see that last way of writing it corresponds to first shifting $3$ to the right then rotating?

19. Mar 28, 2014

### LCKurtz

Good job. I think you have made good progress from where you started.

Last edited: Mar 28, 2014
20. Mar 29, 2014

### skrat

Thanks to you and your paitence.

for e) part, where $Re(z)<-3$ is mapped into open unit circle ... The idea is to map all the vertical lines inside the unit circle, meaning, lines close to $\infty$ will be mapped to origin and lines at $-3$ will be mapped close to the edge of unit circle.

I am again having some troubles here. I have some notes saying that $f(z)=e^z$ maps vertical lines between $|Im(z)|<\pi$ into circles... Let's try to use this.

I want to move the lines into origin, therefore $e^{z+3}$. Let's say that $z=a+it$ where $t\in [-\infty, \infty]$, than $e^{z+3}=e^{a+3+it}=e^{a+3}e^{it}$.

Hmmm, $e^{a+3}$ will do the job, which is map lines from close to infinity to the origin and line at $-3$ to the edge, which is open. $e^{a+3}$ is actually the radius of a circle. However ... I think the second part may be problematic. I know that $e^{2k\pi i}=1$ for $k \in \mathbb{Z}$ but here $t \in \mathbb{R}$.

The way I understand it $e^{it}$ is a complex number that only rotates for $t$ around the origin and it's length is always $1$.

What I am worried about is that $t\in [-\pi ,\pi ]$ already does the job, any greater $t$ will only map on top meaning this transformation may not be bijection?