Find analitical function f that maps conformal

[skrat]

Homework Statement

Find analitical function $f$ that maps half-plane $Re(z)<-3$ conformal on:
a) half-plane $Re(z)>0$
b) half-plane $Im(z)>0$
c) open first quadrant
d) open second quadrant
e) open unit circle

Homework Equations

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): ...?

 

[LCKurtz]

Homework Statement

Find analitical function $f$ that maps half-plane $Re(z)<-3$ conformal on:
a) half-plane $Re(z)>0$
b) half-plane $Im(z)>0$
c) open first quadrant
d) open second quadrant
e) open unit circle

Homework Equations

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): ...?

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

 

[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.

 

[LCKurtz]

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.

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.

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

 

[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"...?

 

[LCKurtz]

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"...?

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.

 

[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.

 

[LCKurtz]

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

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.

 

[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?

 

[LCKurtz]

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

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

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

 

[skrat]

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

 

[skrat]

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

 

[LCKurtz]

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

 

[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.

Huh... Please say yes. :D

 

[LCKurtz]

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.

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

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

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.

 

[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!

 

[skrat]

From $Re(z)<-3$ to
c) open first quadrant
d) open second quadrant

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.

 

[LCKurtz]

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.

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?

 

[LCKurtz]

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.

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

 

[skrat]

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

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?

 

[LCKurtz]

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?

I agree it wouldn't be a bijection. A map of a half plane to the unit circle can be done as a rotation of the Riemann sphere. Have you studied linear fractional transformations? I think that's where you need to look. Many texts do that as an example.

[Edit] Maybe what you have is what is expected. Does it have to be 1-1?

 

[skrat]

I have studied linear fractional transformations. Similar example is actually written in my notes saying that $f(z)=\frac{1-z}{z+1}$ will map $U=\begin{Bmatrix} z; |z|<1 \end{Bmatrix}$ (obviously $U$ is inside of an open unit circle) into right half plane.

I could probably work something out using this ... Firstly, I need left half plane therefore an extra sign will be needed also I want to translate the lines for three units to the right.

EDIT: The solution was clearly wrong... Will be back soon with a better one...

 

[LCKurtz]

I have studied linear fractional transformations. Similar example is actually written in my notes saying that $f(z)=\frac{1-z}{z+1}$ will map $U=\begin{Bmatrix} z; |z|<1 \end{Bmatrix}$ (obviously $U$ is inside of an open unit circle) into right half plane.

I could probably work something out using this ... Firstly, I need left half plane therefore an extra sign will be needed also I want to translate the lines for three units to the right.

$f(z)=\frac{z+2}{z+4}$ yet I need inverse function of that, which gives me $f(z)=\frac{2-z4}{z-1}$.

Or maybe not..?

OK, if you have $f(z)=\frac{1-z}{z+1}$ maps the open circle to the right half plane, then its inverse $g$ would map the half plane to the circle. Since you had earlier that $-z-3$ maps your beginning region to the right half plane, you can apply apply $g$ to $-z-3$ and that should do it. But I don't get the same answer as you have above. Show me what you get for the inverse of $f(z)$ and apply it to $-z-3$.

 

[skrat]

[Edit] Maybe what you have is what is expected. Does it have to be 1-1?

1-1 means bijection?

Well.. It says here that the transformation has to be analytic and it should also map $Re(z)<-3$ conformal. It says nothing about bijection, if transformation is locally conformal it is also locally bijective. Or it is not?

 

[skrat]

I get the same expression for inverse function so $f(z)=f^{-1}(z)=\frac{1-z}{z+1}$.

Now applying $-z-3$ gives me $f(z)=\frac{z+4}{-z-2}$.

BTW: There was no way you could ever get the same answer as I did :D because I first used $-z-3$ (with wrong signs) and than calculated the inverse.

 

[LCKurtz]

That looks good. When you think about it geometrically, $f$ should equal $f^{-1}$ as you found. It represents a rotation half way around of the Riemann sphere and doing it again gets you back where you were. So I guess we are done with this thread. Good work, I think you learned something.