Find all linear transformations which

[brkomir]

Homework Statement

Find all linear transformations $f(z)=az+b$ which map half-plane $Im(z)>0$ on $Im(z)>0$. It is a so called self-mapping transformation.

Homework Equations

The Attempt at a Solution

I am guessing this will have something to do with Möbius transformation, but sadly I haven't got even a clue on how to start with this problem.
If one could please give me hint, I would be really happy.

Cheers,
brko

 

[pasmith]

Homework Statement

Find all linear transformations $f(z)=az+b$ which map half-plane $Im(z)>0$ on $Im(z)>0$. It is a so called self-mapping transformation.

Homework Equations

The Attempt at a Solution

I am guessing this will have something to do with Möbius transformation, but sadly I haven't got even a clue on how to start with this problem.
If one could please give me hint, I would be really happy.

Cheers,
brko

You want linear transformations, and you want $a$ and $b$ such that
$$\mathrm{Im}(az + b) > 0$$
for all $z$ with $\mathrm{Im}(z) > 0$.

 

[brkomir]

Yes. So?I am supposed to work something out from $f(x+y)=f(x)+f(y)$ and $f(\alpha x)=\alpha f(x)$?

 

[Fredrik]

I don't think "linear" means what it normally does in this problem. Note that the b in the problem statement ensures that f isn't linear in that sense.

I would interpret the question like this: For what values of a and b does the function f defined by f(z)=az+b have the property that I am f(z)>0 for all z such that I am z>0?

 

[brkomir]

Hmmm...

Let's say that $z=x+iy$, $a=a_1+ia_2$ and $b=b_1+ib_2$ for $x,y,a_i,b_i \in \mathbb{R}$.

Than $f(z)=(a_1+ia_2)(x+iy)+b_1+ib_2=a_1x+ia_1y+ia_2x-a_2y+b_1+ib_2=a_1x-a_2y+b_1+i(a_1y+a_2x+b_2)$.

Now $Im(f(z))=a_1y+a_2x+b_2>0$ where we already know that $Im(z)=y>0$.

Now I could distinguish cases where $a,b\in \mathbb{R}$ ... Is that a step on a right path?

 

[Fredrik]

That's how I would start the problem too. In the next step, I would use that the inequality you found has to hold for arbitrary choices of x and y. (y must be chosen positive, but x can be chosen to be any real number). So you can ask yourself things like "Is there a choice of x that makes this inequality false?".

 

[brkomir]

$a_1y+a_2x+b_2>0$

The $x>-\frac{b_2+a_1y}{a_2}$. So $x$ can not be $x\leq -\frac{b_2+a_1y}{a_2}$.

But, how does this answer:

I would interpret the question like this: For what values of a and b does the function f defined by f(z)=az+b have the property that I am f(z)>0 for all z such that I am z>0?

I could also say that $a_2 \neq 0$ meaning $a \in \mathbb{C}$ for sure!

 

[Fredrik]

I was thinking that since the inequality is supposed to hold for all $x,y\in\mathbb R$ such that $y>0$, you can make several arguments like this:

Since the inequality has to hold when y=1 and x=0, we must have $a_1+b_2>0$.

Each choice of x and y narrows down the possible values of a and b. I haven't followed this through to the end, so I don't know exactly what the result will be, but this seems like a reasonable way to proceed.

 

[brkomir]

Here is what I came up with:

$a_1y+a_2x+b_2>0$ and therefore $x>\frac{b_2+a_1y}{a_2}$.

Now we already know that $y>0$, let's check everything for $x$.

1.) $x=0$:

$0>\frac{b_2+a_1y}{a_2}$ therefore $b_2<-a_1y$.

2.) $x>0$:

$\frac{b_2+a_1y}{a_2}<0$ therefore $b_2<-a_1y$.

3.) $x<0$:

$\frac{b_2+a_1y}{a_2}>0$ therefore $b_2>-a_1y$.To sum up:

$y=0$, $a_2 \neq 0$, $b_1$ arbitrary. For $a_1, b_2$ we know that if:

1.) $x\geq 0$:

$b_2\leq \left\{\begin{matrix} a_1y;a_1<0\\ -a_1y;a_1>0\\ 0;a_1=0 \end{matrix}\right.$

which can also be written as $b_2\leq |a_1y|$

2.) $x<0$:

$b_2> \left\{\begin{matrix} a_1y;a_1<0\\ -a_1y;a_1>0\\ 0;a_1=0 \end{matrix}\right.$

Which can also be written as $b_2>|a_1y|$If that is all correct, than I really don't see what else could be done here.

 

[Fredrik]

Here is what I came up with:

$a_1y+a_2x+b_2>0$ and therefore $x>\frac{b_2+a_1y}{a_2}$.

Isn't there a sign error here? Or did you determine that $a_2<0$ before this? Since you need to flip the direction of the < symbol when you multiply the inequality by something negative, you probably shouldn't multiply it by something that may or may not be negative.

Now we already know that $y>0$, let's check everything for $x$.

1.) $x=0$:

$0>\frac{b_2+a_1y}{a_2}$ therefore $b_2<-a_1y$.

By setting x=0 in $a_1y+a_2x+b_2>0$, we get $b_2>-a_1y$. But you don't have to end the argument at this point, because there are values of $a_1$ such that this can't possibly be true for all y>0.

 

[brkomir]

Isn't there a sign error here? Or did you determine that $a_2<0$ before this? Since you need to flip the direction of the < symbol when you multiply the inequality by something negative, you probably shouldn't multiply it by something that may or may not be negative.

Ooops, I have the right sign in my papers. Sorry for that. Should be ok everywhere else. Anyway, good point.

By setting x=0 in $a_1y+a_2x+b_2>0$, we get $b_2>-a_1y$. But you don't have to end the argument at this point, because there are values of $a_1$ such that this can't possibly be true for all y>0.

I will get thousands of conditions...

So for $x=0$ than $b_2+a_1y>0$ now I know nothing about $b_2$ or $a_1$ except the fact that they are both real numbers.

$b_2>-a_1y$ gives me positive $b_2$ if $a_1$ is negative and negative $b_2$ if $a_1$ is positive.

Using this knowledge I get that for $x>0$, $a_2<0$ and for $x<0$, $a_2>0$. That is if $a_1 \in \mathbb {R} /(0)$.

 

[Fredrik]

a and b do not depend on x or y. So results like "for $x>0$, $a_2<0$ and for $x<0$, $a_2>0$" can't be correct.

You have found that for all $y>0$, we have $b_2<-a_1y$. This implies among other things that $a_1\leq 0$. (Assume that $a_1>0$ and derive a contradiction).

I'm going to bed, so no more posts from me for 8 hours or so. (Edit: I made a sign error in my first version of this post. Should be OK now).

Edit 2: The inequality I wrote as $b_2<-a_1y$ should be $b_2>-a_1y$. See the correction below.

 

[epenguin]

er well, nonmathematician but couldn't resist, have made fool of self in math before, but I understand from things you say a, b are supposed to be real (though I don't think even that would change the answer) isn't this everything that in the Argand diagram maps points in the top half to points in the top half and isn't it obvious that a ≥ 0, b ≥ 0, a, b not both 0 does this, and anything different will fail to do this for whole areas though not necessarily for everything?

 

[Fredrik]

(I couldn't sleep...will try again in a while). The problem doesn't explicitly say what a and b are, but I think it's safe to assume that we're supposed to find all complex values of those variables that ensure that give f the desired property. That's why we write $a=a_1+ia_2$ and similarly for b.

If we had been looking only for real values, then the inequality $a_1y+a_2x+b_2>0$ would have been simply $a_1y>0$ (because $a_2=b_2=0$). This holds for all y>0. So if $a_1\neq 0$, we have $a_1|a_1|>0$, which implies $a_1>0$. So we have $a_1\geq 0$. There would be no restrictions on b.

 

[brkomir]

(I couldn't sleep...will try again in a while).

I'm so sorry! :D

However, since this everything is an introduction to complex analysis, I think it is safe to assume that $a,b \mathbb{C}$.

You have found that for all $y>0$, we have $b_2<-a_1y$. This implies among other things that $a_1\leq 0$. (Assume that $a_1>0$ and derive a contradiction).

I think there should be$b_2>-a_1y$.
How does this imply anything? $b_2$ can be either positive either negative and the same goes for $a_1$.
In other words: if $a_1$ is negative than $b_2$ will be positive.

Or am I making fun of myself here?

 

[Fredrik]

Ah, yes, the result we got is that for all y>0, $b_2>-a_1y$. (This sort of thing happens a lot when I post just before I go to bed). This implies among other things that $a_1\geq 0$, because if $a_1<0$ you can choose y so that the inequality $b_2>-a_1y=|a_1|y$ is violated. No matter what the value of $b_2$ is, you can choose y so that $|a_1|y$ is greater than that.

 

[brkomir]

Ok, makes sense.

$a_1y+a_2x+b_2>0$ and now we know that $y>0$ also $a_1\geq 0$ therefore $a_1y \geq 0$.

$b_2>-a_1y-a_2x$ where if $a_2<0$ I can find such $x$ that the inequality is violated. The same goes if $a_2>0$. Does this mean that $a_2=0$?

 

[epenguin]

If we had been looking only for real values, then the inequality $a_1y+a_2x+b_2>0$ would have been simply $a_1y>0$ (because $a_2=b_2=0$). This holds for all y>0. So if $a_1\neq 0$, we have $a_1|a_1|>0$, which implies $a_1>0$. So we have $a_1\geq 0$. There would be no restrictions on b.

Right, any real b just maps everything sideways so no restrictions, I did think that but later lost sight of it , it was late night for me too. . That leaves a ≥ 0 for real a, and still simple.

For complex a, b I would be thinking along lines that in general (i.e. for nonreal complex numbers) the transformation would involve a rotation about some point and you cannot nontrivially rotate an infinite half-plane about any point, even combining with any translation without some of it ending up in the other (Im(z) < 0) half plane.

 

[Fredrik]

Ok, makes sense.

$a_1y+a_2x+b_2>0$ and now we know that $y>0$ also $a_1\geq 0$ therefore $a_1y \geq 0$.

$b_2>-a_1y-a_2x$ where if $a_2<0$ I can find such $x$ that the inequality is violated. The same goes if $a_2>0$. Does this mean that $a_2=0$?

Don't abandon the result "For all y>0, we have $b_2>-a_1y$" yet. It also tells you something about $b_2$.

I believe that you're right about $a_2$. If $a_2\neq 0$, we can give the term $a_2x$ any real value we want, and that's a problem.

 

[brkomir]

Forgot to write it..

Of course, $b_2>-a_1y$ so $b_2\leq 0$

 

[Fredrik]

Actually it's $b_2\geq 0$. But we're close to the finish now. We have found that $a_1,b_2\geq 0$ and $a_2=0$. However, we have only proved that if f has the desired properties, then a and b must satisfy the results we found. Now you need to think about whether it's also the case that if a and b satisfy those results, then f has the desired property.

 

[brkomir]

Actually it's $b_2\geq 0$. But we're close to the finish now.

No, it is not. ?!

$y>0$ also $a_1\geq 0$. The product therefore $a_1y \geq 0$. Aha. Ok. I take that back. It is $b_2 \geq 0$.

Now you need to think about whether it's also the case that if a and b satisfy those results, then f has the desired property.

That was our goal right? To find such a and b that f will have the desired property. And we found them. Why would it be any different the other way around?

 

[Fredrik]

Because logically $p\Rightarrow q$ isn't equivalent to $q\Rightarrow p$. If you prove that $p\Rightarrow q$, it's still possible that the implication $q\Rightarrow p$ is false, i.e. that q is true and p is false. For example, x=1 implies that x(x-1)=0. But x(x-1)=0 doesn't imply that x=1.

In our case, p is the statement that f takes the upper half-plane to itself, and q is the statement that all of the following are true: $a_1\geq 0$, $b_2\geq 0$, $a_2=0$.

It's not hard to see that it's not the case that all choices of a and b that satisfy these condition ensure that f has the desired property.

 

[Fredrik]

This is probably a better explanation: Each choice of values of x and y restricts the possible values of a and b. For example, if we chose x=y=1, we see that $a_1+a_2+b_2>0$. This tells us that the pair (a,b) is in the set $\{(a,b)\in\mathbb C^2|a_1+a_2+b_2>0\}$. Each choice of x and y gives us a subset of $\mathbb C^2$ that (a,b) must be in. The result we got, $a_1\geq 0, b_2\geq 0, a_2=0$, defines a subset of $\mathbb C^2$ too. This set is the intersection of the subsets corresponding to the values of x and y that we tried. But we didn't try all of them.