Complex numbers again

**bobak** · April 13th 2008, 03:57 PM

The transformation from the z plane to the w plane is given by

$w = \frac{z-i}{z}$

i) show that under this transformation the line $Im(z) = \frac{1}{2}$ is mapped to the circle with equation $|w| = 1$

ii) hence or otherwise find in the form $w = \frac{az + b}{cz+d}$ where a ,b ,c and d $\in \mathbb{C}$ the transformation that maps the line $Im(z) = \frac{1}{2}$ to the circle centre $(3 -i)$ and radius 2

part i, I didn't have any problems with, i wrote z in the form of x + 0.5i did the transformation.

part ii is getting me, its keeps getting messy, i assume there is a quick solutions but I have missed it

Bobak

**ThePerfectHacker** · April 13th 2008, 04:40 PM

There is an easy solution to this problem.

Just one minor point (a pun) $z\mapsto (z-i)/z$ does not really map the line $\Im(z) = 1/2$ onto $|z|=1$ . Because there is one point on this circle you will get really close to as $|y|$ gets large but you will never reach. If you want you can say that $z\mapsto (z -i)/z$ maps $\Im (z) = 1/2 \cup \{\infty \}$ onto $|z|=1$ where $\infty$ is the point at infinity on the Riemann sphere. But this is nothing serious, so will not pay attention to this problem.

We know that $f_1(z) = (z-i)/z$ maps $\Im (z) = 1/2$ onto $|z|=1$ . But we also know that $f_2(z) = 2z$ will enlarge the circle $|z|=1$ to $|z|=2$ . Finally, $f_3(z) = z + (3-i)$ will map $|z|=2$ onto $|z-(3-i)| = 2$ .

Thus, $f_3\circ f_2\circ f_1 (z)$ will map $\Im (z) = 1/2$ onto $|z-(3-i)| = 2$ . But, $f_3\circ f_2 \circ f_1 (z) = \left( 2\cdot \frac{z-i}{z} \right) + (3-i)$ . Convert this into the form $(az+b)/(cz+d)$ and you get your answer.

By the way. Here is an interesting theorem from complex analysis. Functions of the form $(az+b)/(cz+d)$ (where $ad-bc\not = 0$ ) are called Mobius Transformation. They have the nice property that they map lines and circles to lines and circles.

**bobak** · April 13th 2008, 05:31 PM

So we have a dilation and a translation right?

i got $\frac{(5-i)z -2i}{z}$ which is the answer in the book

Many thanks TPH

Bobak

**ThePerfectHacker** · April 13th 2008, 06:02 PM

Originally Posted by bobak

So we have a dilation and a translation right

Exactly that is what I did in steps.

Here are some basic transformations of the complex plane.

$f(z) = kz$ , where $k$ is a positive real number. If $k>1$ then the shape gets magnified by a factor of $k$ if $k<1$ then the shape gets minimized by a factor of $1/k$ .
$f(z) = z + \alpha$ where $\alpha$ is a complex number. This is a translation in the direction of $\alpha$ .
$f(z) = e^{i\theta} z$ . Where $\theta$ is a real number. This is a counter-clockwise rotation by $\theta$ .

When I was studing conformal mappings I came up with a very simple function which will make all problems involving lines and circles really easy. It is $f(z) = (1-z)/(1+z)$ .

This function has some really nice properties. Which you might want to try to prove.

$f(z)$ will map $|z|=1$ onto $\Re (z) = 0$ .
$f(z)$ will map $\Re(z) = 0$ onto $|z|=1$ .
$f(z) = f^{-1}(z)$

Actually we should be a little careful. $f(z)$ is not defined at $z=-1$ , at that point on the unit circle. So it is more formal to say $\{ |z| = 1\} - \{ -1\}$ gets mapped to $\Re (z) = 0$ . And conversely, $f(z)$ does not really map $\Re (z) = 0$ to $\{ |z| = 1\}$ but rather to $\{ |z| = 1\} - \{ -1\}$ . However, this is a minor problem which we can fixed like we did now.

For example, suppose we want to find a mapping from $\Im (z) = 1/2$ to $|z| = 1$ . I will state this right now, there are many many such mappings, i.e. what we will find it not unique. However, we can find this mapping without any algebra! Just using the geometric facts above. The first step is to turn this line $\Im (z) = 1/2$ to the line $\Re (z) = 0$ . The reason being is because we know how to deal with the line $\Re (z) = 0$ . (The approach I use is not unique, if you see your own way of doing it, it will also work). First, map $\Im (z) = 1/2$ to $\Im (z) = 0$ , we do this easily by $f_1(z) = z - (1/2)i$ . Second, we rotate the y-axis to the x-axis, we can do this by $f_2(z) = e^{\pi/2 i} z = iz$ . Now we are at $\Re (z) = 0$ . Third, use the function $f_3(z) = (1-z)/(1+z)$ to get the unit circle. Thus, $f_3 \circ f_2 \circ f_1 (z)$ will do the job.

**bobak** · April 14th 2008, 06:23 AM

Originally Posted by ThePerfectHacker

Exactly that is what I did in steps.

Here are some basic transformations of the complex plane.

$f(z) = kz$ , where $k$ is a positive real number. If $k>1$ then the shape gets magnified by a factor of $k$ if $k<1$ then the shape gets minimized by a factor of $1/k$ .
$f(z) = z + \alpha$ where $\alpha$ is a complex number. This is a translation in the direction of $\alpha$ .
$f(z) = e^{i\theta} z$ . Where $\theta$ is a real number. This is a counter-clockwise rotation by $\theta$ .

When I was studing conformal mappings I came up with a very simple function which will make all problems involving lines and circles really easy. It is $f(z) = (1-z)/(1+z)$ .

This function has some really nice properties. Which you might want to try to prove.

$f(z)$ will map $|z|=1$ onto $\Re (z) = 0$ .
$f(z)$ will map $\Re(z) = 0$ onto $|z|=1$ .
$f(z) = f^{-1}(z)$

Actually we should be a little careful. $f(z)$ is not defined at $z=-1$ , at that point on the unit circle. So it is more formal to say $\{ |z| = 1\} - \{ -1\}$ gets mapped to $\Re (z) = 0$ . And conversely, $f(z)$ does not really map $\Re (z) = 0$ to $\{ |z| = 1\}$ but rather to $\{ |z| = 1\} - \{ -1\}$ . However, this is a minor problem which we can fixed like we did now.

I managed all of that okay.

For example, suppose we want to find a mapping from $\Im (z) = 1/2$ to $|z| = 1$ . I will state this right now, there are many many such mappings, i.e. what we will find it not unique. However, we can find this mapping without any algebra! Just using the geometric facts above. The first step is to turn this line $\Im (z) = 1/2$ to the line $\Re (z) = 0$ . The reason being is because we know how to deal with the line $\Re (z) = 0$ . (The approach I use is not unique, if you see your own way of doing it, it will also work). First, map $\Im (z) = 1/2$ to $\Im (z) = 0$ , we do this easily by $f_1(z) = z - (1/2)i$ . Second, we rotate the y-axis to the x-axis, we can do this by $f_2(z) = e^{\pi/2 i} z = iz$ . Now we are at $\Re (z) = 0$ . Third, use the function $f_3(z) = (1-z)/(1+z)$ to get the unit circle. Thus, $f_3 \circ f_2 \circ f_1 (z)$ will do the job.

Okay I gave this a shot, I got the function as

$f(z) = \frac{1-2iz}{3+2iz}$ and did a check to see if it did the mapping correctly. so this does the same transformation on $\Im(z) = \frac{1}{2}$ mapping as $f(z) = \frac{z-i}{z}$ .

Out of interest why are the mappings not unique?

Many thanks for taking your time to explain all of this to me TPH, I really appreciate it

.

Bobak

**ThePerfectHacker** · April 14th 2008, 08:42 AM

Originally Posted by bobak

Out of interest why are the mappings not unique?

There are infinitely many of them. That has to do with the fact you can have different functions which map the same lines into the same circles. While what these functions do to other points, line, and circles is different. In general given a triple of (distinct) complex points $(a,b,c)$ you can find a Mobius transformation which sent those to $(d,e,f)$ (in that order). Since lines are determined by two points and circles by three (non-collinear) point this explains why are so many. Meaning if lines where determined by three point then yes the mapping will be unique by the properties of Mobius transformations, but the problem is lines are not determined by three points.
---
If you are willing to work on a more difficult problem you can try to prove that $f(z) = (1-z)/(1+z)$ maps the open disk $\text{D}(0;1)$ * bijectively onto the right half-plane i.e. $\{ z\in \mathbb{C} : \Re (z) > 0 \}$ .

I am not sure how easy or hard this problem is. I never tried proving it, just always accepted it on faith. It does not seem hard.

*)This notation, $\text{D}(0;1)$ is the region $|z| < 1$ , i.e. disk at $0$ with radius $1$ . We call it a "disk" rather than a "circle" because it contains all the stuff inside too. And we call this to be the "open disk" because for the same reason we call the interval $(0,1)$ "open" because it does not contain its endpoints. Here the endpoints is the boundary of the disk, i.e. $|z|=1$ .

Do you know what "bijectively" means? I am not sure if you ever seen this word before, I know you are still in high-school. As you learn math further you will constantly see this word. It means the mapping is one-to-one, i.e. if $f(z_1) = f(z_2) \implies z_1 = z_2$ , i.e. distinct points get mapped to distinct points. And it also means any point $w$ you choose in the right half plane can be obtained from $f(z) = w$ where $z$ is in the open unit disk.

**bobak** · April 14th 2008, 09:36 AM

I half done it.

we have a transformation form the z plane to the w plane defined by $w = \frac{1-z}{1+z}<br />$ the transformation maps points form z (where $z = x+iy$ ) onto the point $\frac{1-(x^2+y^2)}{(1+x)^2+y^2} - \frac{2y}{(1+x)^2+y^2}i$ in w plane.

$\forall,x,y \in \text{D}(0;1) \ \ x^2+y^2 < 1 \Rightarrow \Re(w) > 0<br />$ (is this correct notation ?).

I am unsure on how to show that this is bijective. I know that for a fixed x in the z plane all the imaginary part of the point in the w plane are distinct, however is far form complete, could you give me a pointer ?

Many Thanks

Bobak

**ThePerfectHacker** · April 14th 2008, 10:11 AM

Originally Posted by bobak

(is this correct notation ?).

Yes.

I am unsure on how to show that this is bijective. I know that for a fixed x in the z plane all the imaginary part of the point in the w plane are distinct, however is far form complete, could you give me a pointer ?

Remember what you need to show.

$f(z)$ maps disk to point in right half-plane
For any point $w$ in right half-plane there exists a $z$ in the unit disk so that $f(z) = w$
$f(z)$ is one-to-one

So far you proved #1. If you look at #3 it happens to be easy to show. The most difficult one is #2.

I am going to give you a hint on #2. Avoid mindless algebra, meaning do not just let $w=x+iy$ and try to solve system of equations for $x,y$ . Look for elegance. I did it on paper and it was not messy at all, if you are doing it carefully.

You proved #1 correctly. But here is an elegant way to prove it. Note, there is no mindless algebra going on over here.

We want to show if $|z|<1$ then $\Re \left( \frac{1 - z}{1+z} \right) > 0$ .
Note, $\Re \left( \frac{1-z}{1+z} \right) = \Re \left( \frac{1-z}{1+z}\cdot \frac{\overline{1+z}}{\overline{1+z}}\right) = \Re \left( \frac{1-z}{1+z} \cdot \frac{1+\bar z}{1 + \bar z}\right)$

But $\Re \left( \frac{\alpha}{r} \right) > 0$ if and only if $\Re (\alpha) > 0$ (here $\alpha$ is complex and $r>0$ ).

Thus, $\Re \left( \frac{1-z}{1+z} \cdot \frac{1+\bar z}{1 + \bar z}\right) > 0$ if and only if $\Re [(1- z)(1+\bar z)] > 0$ .
But, $(1-z)(1+\bar z) = 1 - (z-\bar z) - z\bar z = (1 - |z|^2) - 2i \Im (z)$
Thus, $\Re [(1-z)(1+\bar z)] = 1 - |z|^2 > 0$ because $|z|<1$ .

**bobak** · April 14th 2008, 11:32 AM

okay for 2.

We know that $f(z)$ is its own inverse so $f(w)$ will map points form the w plane back onto the z plane.

so we must show that for $\Re(w) > 0 \ \ \left| \frac{1-w}{1+w} \right| < 1$
$\Rightarrow |1-w| < |1+w|$ which is clearly true as the real part of the right hand side is always greater than the real part of the left and side and the imaginary he parts are equal.

is that okay ?

I am stuck on 3 though

Bobak

**ThePerfectHacker** · April 14th 2008, 03:38 PM

Originally Posted by bobak

so we must show that for $\Re(w) > 0 \ \ \left| \frac{1-w}{1+w} \right| < 1$
$\Rightarrow |1-w| < |1+w|$ which is clearly true as the real part of the right hand side is always greater than the real part of the left and side and the imaginary he parts are equal.

is that okay ?

Why? How do you know that $|1-w| < |1+w|$ just because $\Re (w) > 0$ ?

I was doing this problem differently.
First,
$|1-w| < |1+w|$ iff (if and only if)
$|1-w|^2 < |1+w|^2 \implies (1-w)(\overline{1-w}) < (1+w)(\overline{1+w})$ $\implies (1-w)(1-\bar w)<(1+w)(1+\bar w)$
Thus, $1 - w - \bar w + w\bar w < 1 + w+\bar w + w\bar w\implies -(w+\bar w) < w+\bar w \implies - 2\Re (w) < 2\Re (w)$
This is of course true because $\Re(w) > 0$ .

I am stuck on 3 though

There is not much to show on #3. Because we shown that $f^{-1}$ exists (and furthermore is equal to itself). The inverse function only exists if the function is one-to-one.

But if you want to do it differently it is also easy.
Say $(1-z_1)/(1+z_1) = (1-z_2)/(1+z_2) \implies 1 - z_1 + z_2 - z_1z_2 = 1 - z_2 + z_1 - z_1z_2$ $\implies 2z_1 = 2z_2 \implies z_1 = z_2$ .

Here are two related problems to try.
1)Show that $f(z) = (az+b)/(cz+d)$ is a one-to-one function if $ad-bc \not = 0$ .
2)Find a mapping* from $\text{D}(0,2)$ to the upper half plane.

*)The appropriate term is "conformal mapping". A conformal mapping is a one-to-one analytic function. The word "analytic" in complex analysis simply means differenciable. So if you can find a derivative then it is analytic. If this is confusing you just ignore all these terms.

**bobak** · April 14th 2008, 03:48 PM

Originally Posted by ThePerfectHacker

Why? How do you know that $|1-w| < |1+w|$ just because $\Re (w) > 0$ ?

well because $1+x > 1-x \ \ \forall x >0$ so the length of he real part is always greater and the imaginary part is fixed. surely that is good enough ?

There is not much to show on #3. Because we shown that $f^{-1}$ exists (and furthermore is equal to itself). The inverse function only exists if the function is one-to-one.

Of course

, i had a funny feeling that i had to use the inverse properly but i wasn't too sure about it.

Here are two related problems to try.
1)Show that $f(z) = (az+b)/(cz+d)$ is a one-to-one function if $ad-bc \not = 0$ .
2)Find a mapping* from $\text{D}(0,2)$ to the upper half plane.

*)The appropriate term is "conformal mapping". A conformal mapping is a one-to-one analytic function. The word "analytic" in complex analysis simply means differenciable. So if you can find a derivative then it is analytic. If this is confusing you just ignore all these terms.

I will try this tomorrow, its late over here.

Thank again for your help TPH

Bobak

**ThePerfectHacker** · April 15th 2008, 12:15 PM

These functions that we have been looking at, for example, $f(z) = (1-z)/(1+z)$ are called conformal mappings. So you probably thinking why the word "conformal"? That word means "angle preserving". And this function has an special property that it preserves angles of curves. First, we define an angle of two intersecting curves (not lines) to be the angle between their tangents at that point. I am claiming that if you have a pair of curves intersecting at an angle then the images of those curves will intersect at the same angle under the function $f(z)$ .

Here are some examples that you might want to try.

1)Show that the lines $x=0$ and $y=0$ which are perpendicular get mapped into perpendicular (orthogonal) circles under $f(z)$ .

2)Let $g(z) = 1/z$ (this is called an "inversion transformation"). Show that all the lines $x=\pm 1,\pm 2$ and $y=\pm 1,\pm 2$ get mapped into orthogonal circles. (The reason why we are avoiding the lines $x=0$ and $y=0$ is because the function would be undefined at the origin, so let us just avoid that).

3)After you get convined that $g(z)$ is a conformal map, i.e. it preserves angles. You can go here here. These pictures are showing what the conformal functions are doing to the x-y gird. Meaning you draw an integer grid ( $x=k,y=m$ where $k,m$ integers you will get an integer grid, all curves are perpendicular) you will get orthogonal curves after you apply each of those functions in the list.

But not everything is conformal. For example, $f(z) = z^2$ is conformal everywhere except for $z=0$ . Because at the origin there are two curves which get mapped into different angles. Try to find and example of those curves.

So a natural question to ask: Is there a way to determine whether a complex-valued function is conformal? And the answer is yes. In fact, it is a very simple and very supprising answer.

**bobak** · April 16th 2008, 03:41 AM

Originally Posted by ThePerfectHacker

Here are two related problems to try.
1)Show that $f(z) = (az+b)/(cz+d)$ is a one-to-one function if $ad-bc \not = 0$ .
2)Find a mapping* from $\text{D}(0,2)$ to the upper half plane.

Not sure how neat this is.

For 1)

I let $f(z_1) = f(z_2)$

$\frac{az_1+b}{cz_1+d}=\frac{az_2+b}{cz_2+d}$

... (omitted algebra)

$z_1(ad-bc) = z_2(ad-bc)$

so $z_1 = z_2$ providing $ad-bc \neq 0$

so the function is bijective.

for 2)

We know that $\frac{1-z}{1+z}$ maps $\text{D}(0,1)<br />$ to the region in the w plane $\Re(w) > 0$

and now require a function to map $\text{D}(0,2)<br />$ onto $\Im(w) > 0$

so I define three functions

$f_1(z) = 2z$ to increase the modulus of the unit disc
$f_2(z) = \frac{1-z}{1+z}$ to map the points onto $\Re(w) > 0$
$f_3(z) = e^{\frac{\pi}{2}i}z$ to rotate, changing $\Re(w) > 0$ to $\Im(w) > 0$

so the mapping is $f_3\circ f_2\circ f_1 (z) = \frac{i(1-2z)}{1+2z}$

Is this all good?

many thanks

Bobak

**ThePerfectHacker** · April 16th 2008, 06:49 AM

Originally Posted by bobak

I let $f(z_1) = f(z_2)$

$\frac{az_1+b}{cz_1+d}=\frac{az_2+b}{cz_2+d}$

... (omitted algebra)

$z_1(ad-bc) = z_2(ad-bc)$

so $z_1 = z_2$ providing $ad-bc \neq 0$

so the function is bijective.

You did not actually prove it is bijective. You proved it is one-to-one. And that is exactly how it is done.

$f_1(z) = 2z$ to increase the modulus of the unit disc
$f_2(z) = \frac{1-z}{1+z}$ to map the points onto $\Re(w) > 0$
$f_3(z) = e^{\frac{\pi}{2}i}z$ to rotate, changing $\Re(w) > 0$ to $\Im(w) > 0$

so the mapping is $f_3\circ f_2\circ f_1 (z) = \frac{i(1-2z)}{1+2z}$

It is good except you made a minor mistake, $f_1(z) = (1/2)z$ . You want to shrink $\text{D}(0,2)$ to $\text{D}(0,1)$ . But everything else is perfect.

**bobak** · April 17th 2008, 02:00 AM

Originally Posted by ThePerfectHacker

1)Show that the lines $x=0$ and $y=0$ which are perpendicular get mapped into perpendicular (orthogonal) circles under $f(z)$ .

I am having really difficult determining how the line y = 0 get mapped under this transformation, Well i don't actually know what I am doing, I just do the whole $z = x +i0$ so $w = \frac{1-x}{1+x}$ is that even a circle ?

Some guidance would be appreciated.

Many thanks.

Bobak

Thread: Complex numbers again

LinkBack

Thread Tools

Display

Complex numbers again

Similar Math Help Forum Discussions

raising complex numbers to complex exponents?

Finding real numbers in equations containing complex numbers

What is complex slope of a line in complex numbers

Pre-Calc Graphing with Complex numbers. Multiplying with complex numbers.

Complex Numbers- Imaginary numbers

Search Tags