Complex numbers again

**ThePerfectHacker** · April 17th 2008, 07:43 AM

Originally Posted by bobak

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 ?

The line $y=0$ is the set of points $L=\{ t + 0i|t\in \mathbb{R} \}$ . This means $f(L) = \left\{ \frac{1+t}{1-t}+0i: t\in \mathbb{R} \right\}$ . As $t$ ranges over the real numbers $(1-t)/(1+t)$ ranges over the real numbers (if you want you can add that $t\not = -1$ ). So $f(L) = \{r + i0|r\in \mathbb{R}\}$ which is still the line $y=0$ . Thus, the line $y=0$ gets mapped into itself. If you do the same thing with the line $x=0$ I think you will see it gets mapped into itself. So the angles are definitely preserved here.

That was not a very interesting problem. The other ones are more interesting.

**bobak** · April 20th 2008, 08:51 AM

Originally Posted by ThePerfectHacker

The line $y=0$ is the set of points $L=\{ t + 0i|t\in \mathbb{R} \}$ . This means $f(L) = \left\{ \frac{1+t}{1-t}+0i: t\in \mathbb{R} \right\}$ . As $t$ ranges over the real numbers $(1-t)/(1+t)$ ranges over the real numbers (if you want you can add that $t\not = -1$ ). So $f(L) = \{r + i0|r\in \mathbb{R}\}$ which is still the line $y=0$ . Thus, the line $y=0$ gets mapped into itself. If you do the same thing with the line $x=0$ I think you will see it gets mapped into itself. So the angles are definitely preserved here.

That was not a very interesting problem. The other ones are more interesting.

maybe i got this wrong the line $y=0$ is mapped onto the circle $|z|=1$ and $x=0$ is mapped onto it self to they intersect at right angles at the point $(-1,0)$

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).

I am not really happy with how i am doing this.

for $x = \pm 1$ I get $w= \frac{\pm1}{1+y^2} -\frac{y}{1+y^2}$ we some messy algebra and guess work. i pulled out $\left| w\pm\frac{1}{2} \right| = \frac{1}{2}$

for the lines $y = \pm 1$ $w = \frac{x}{x^2+1} \pm \frac{1}{x^2+1}i$ so we have the the circles with equations $\left| w\pm\frac{1}{2}i \right| = \frac{1}{2}$

so all the circles intersect at right angles at the origin, so the lines are mapped onto orthogonal circles.

But is there a better way to get these circle definitions out other than just my messy algebra? could you show me the elegant way please

many thanks

Bobak

**ThePerfectHacker** · April 21st 2008, 05:54 PM

Originally Posted by bobak

But is there a better way to get these circle definitions out other than just my messy algebra? could you show me the elegant way please

You did it the correct way, I just wanted you to get convinced that the map is angle perserving like I said. There happens to be an cheap way of doing this, that is to use the theorem: Mobius transformations map lines and circles into line and circles. Since $1/z$ if a special case of the Mobius transformation it maps lines into circles or lines. So just pick three random points on the line and map them under $1/z$ if they are collinear then you get a line, if not, then you get the unique circle which passes through those points. The proof of theorem that Mobius transformations map lines and circles into line and circles can be done with mindless algebra again, i.e. substitute $x+iy$ into $(az+b)/(cz+d)$ and get messy. But there is a nice way to proving it, I will post it later.

If you remember I told you that many functions are conformal (angle preserving). There is a simple theorem in complex analysis which tells us which functions are conformal. If $f'(z_0) \not = 0$ for a point $z_0$ then it is conformal there. Meaning if $c_1,c_2$ are curves which intersect at $z_0$ then $f(c_1)$ and $f(c_2)$ are the same angle. Look how simple!

For example, $f(z) = e^z$ is conformal at any point. Because $f'(z_0) = e^{z_0} \not = 0$ . Try it! Pick a few perpendicular lines emanating from the origin and find where they get mapped to. In particular try the easy ones first, i.e. $x=0$ and $y=0$ (which happen to be perpendicular).

Consider, $f(z) = z^2$ , this function is conformal everywhere except possibly at $z_0 = 0$ . Show that at $z_0 = 0$ the function is not conformal by finding a counterexample.

Finally, find where the integer grid (except the line $x=0$ and $y=0$ because they pass through the origin) gets mapped to. For instance, the line $x=1$ gets mapped into a parabola. In this problem, the algebra will get messy, so I suggest to use a graph program so graph these grid lines under $z^2$ and show they are orthogonal.

**ThePerfectHacker** · April 24th 2008, 07:36 PM

You never responded to this. Maybe you did not see it. I can keep on going on this mapping stuff unless it is starting to get boring to you.

**bobak** · April 26th 2008, 10:26 AM

Sorry for the late reply i was going to reply earlier but got side-tracked.

Originally Posted by ThePerfectHacker

You did it the correct way, I just wanted you to get convinced that the map is angle perserving like I said. There happens to be an cheap way of doing this, that is to use the theorem: Mobius transformations map lines and circles into line and circles. Since $1/z$ if a special case of the Mobius transformation it maps lines into circles or lines. So just pick three random points on the line and map them under $1/z$ if they are collinear then you get a line, if not, then you get the unique circle which passes through those points. The proof of theorem that Mobius transformations map lines and circles into line and circles can be done with mindless algebra again, i.e. substitute $x+iy$ into $(az+b)/(cz+d)$ and get messy. But there is a nice way to proving it, I will post it later.

Yeah I figured out how to use my graphing packages parametric equation mode to get the graphs of the transformation, I don't fancy going through the mindless algebraic proof of this theorem. I'll be interested if you post it.

If you remember I told you that many functions are conformal (angle preserving). There is a simple theorem in complex analysis which tells us which functions are conformal. If $f'(z_0) \not = 0$ for a point $z_0$ then it is conformal there. Meaning if $c_1,c_2$ are curves which intersect at $z_0$ then $f(c_1)$ and $f(c_2)$ are the same angle. Look how simple!

Seems almost too simple! What is the theorem called ? and how is it proved? If the proof is very detailed don't bother posting it, I can always purchase a book on the topic, could you recommend a good one though ?

For example, $f(z) = e^z$ is conformal at any point. Because $f'(z_0) = e^{z_0} \not = 0$ . Try it! Pick a few perpendicular lines emanating from the origin and find where they get mapped to. In particular try the easy ones first, i.e. $x=0$ and $y=0$ (which happen to be perpendicular).

this one is pretty easy (providing I got it right)

for $\Im(z) = 0$ $\Im(f(z)) = 0 \ \forall \ z \in \mathbb{C}$ and $\Re(z) > 0$ so it maps the line $y = 0$ onto the half line $\arg( \theta) = 0$

and for $x = 0$ then $f(z) = \sin y + i \cos y$ which is a circle in the complex plane and it intersects with the line at right angles at the point $(1,0)$

Consider, $f(z) = z^2$ , this function is conformal everywhere except possibly at $z_0 = 0$ . Show that at $z_0 = 0$ the function is not conformal by finding a counterexample.

for $z = x + iy$ $f(z) = x^2 - y^2 + 2xyi$

so the line $x=0$ is mapped onto is mapped onto the half line $\arg(\theta) = \pi$ and the line $y=0$ is mapped onto the half line $\arg(\theta) = 0$ . I am not sure if its correct to say that the lines either do not intersect at all at $z_0 = 0$ or if i say they intersect at the angle $\pi$ , nevertheless I have show that function does not preserve angles at $z_0 = 0$ .

Finally, find where the integer grid (except the line $x=0$ and $y=0$ because they pass through the origin) gets mapped to. For instance, the line $x=1$ gets mapped into a parabola. In this problem, the algebra will get messy, so I suggest to use a graph program so graph these grid lines under $z^2$ and show they are orthogonal.

So the integer grid is comprised of the lines $x = \pm a \ \ \forall \ a \in \mathbb{N}$ . so $f(z) = a^2 - y^2 \pm2iay$

so the parabolas are defined parametrically by $y = 2at$ and $x = a^2 -t^2$

lines $y = \pm a \ \ \forall \ a \in \mathbb{N}$

we have parabolas defined by $y = 2at$ and $x = t^2 - a^2$ so these parabola are reflection of the other ones in the y axis.

So the integer grid is transformed onto many intersection parabolas, I attached an image I made using my graphing package.

Now I must show that the tangents of these parabola intersect at right-angles.

for
$y = 2at$ and $x = a^2 -t^2$

$\frac{dy}{dx} = -\frac{a}{t}$

and for $y = 2bt$ and $x = t^2 - b^2$

$\frac{dy}{dx} = \frac{b}{t}$ .

these lines before the intersection would have intersection at the point $(a,b)$ the new points are given by $t = b$ and $t =a$ respectively.

for the product of the gradients on the tangents of these parabola at these points is $-\frac{a}{b} \times \frac{b}{a} = -1$ , hence they are perpendicular, so the function $f(z) = z^2$ is conformal on the interger grid, excluding the lines $y=0$ and $x=0$ .

That was a brilliant exercise, many thanks TPH, sorry again for my late reply I really do enjoy and appreciate all your input.

best regards, Bobak

**ThePerfectHacker** · April 27th 2008, 05:41 PM

That was a brilliant exercise, many thanks TPH, sorry again for my late reply I really do enjoy and appreciate all your input.

You did everything correctly and you made a very nice picture. I gave you this exercise so you will see how these pictures are constructed. If you look at $f(z) = z^2/2$ it will look like your picture except the curves are a little closer because of the presence of the $1/2$ factor.
If you want to try another exercise you can try $f(z)=1/z$ . But there is no point I can see you got the idea.

I don't fancy going through the mindless algebraic proof of this theorem. I'll be interested if you post it.

I will prove to you that if $S$ is a circle or a line and $f(z)$ is a Mobius transformation then $f(S)$ is a circle or a line. The proof is not messy at all and looks nice.

We we call: rotations, translations, dilations, and inversions (that is $1/z$ ) the "basic transformations". The key step is to note that every Mobius transformation is made out of (more formally is a composition of) the "basic transformations".

For example, consider $f(z) = (z+1)/z = 1 + 1/z$ thus $f = f_2 \circ f_1$ where $f_1 = 1/z$ and $f_2=1$ . Another example, $f(z) = (2z-1)/(z+2) = [2(z+2) -5]/(z+2) = 2 - 5/(z+2)$ thus $f =f_4\circ f_3 \circ f_2 \circ f_1$ where $f_1 = z+2$ , $f_2 = 1/z$ , and $f_3 = -5z$ and $f_4 = z+2$ .

The above examples illustrate that you can think of any Mobius transformation as made up the the basic transformations. Note that rotations, dilations, and translations map lines and circles into lines and circles. Thus, to prove that Mobius transformations take lines and circles to lines and circles all we need to show is that the inversion takes lines and circles into lines and circles.

First, we start with a circle. Let $S$ be a set of complex points which is a circle. Let that circle be described as $|z-\alpha| = r$ , where $\alpha$ is its center and $r$ is its radius. Let $f(z) = 1/z$ be the inversion mapping. Then $f(S) = \{ 1/z | z\in S \}$ . We need to show $f(S)$ is a line or a circle.

We know $|z-\alpha| = r$ it means $|z-\alpha|^2 = r^2$ but that means $(z-\alpha)(\overline{z-\alpha}) = r^2$ thus $(z-\alpha)(\bar z - \bar \alpha) = r^2$ . Thus, $|z|^2 - \alpha \bar z - \bar \alpha z = r^2- |\alpha|^2$ . Now if $w\in f(S)$ it means $w = 1/z$ which implies $z = 1/w$ . Thus, if $w\in f(S)$ then the point satisfies $1/|w|^2 - \alpha / \bar w - \bar \alpha / w = r^2 - |\alpha|^2$ .

There are two cases: either $r^2 - |\alpha|^2 = 0$ , i.e. the circle passes through the origin or either $r^2 - |\alpha|^2 \not = 0$ , i.e. the circle does not pass through the origin.

Say that $r^2 - |\alpha|^2 = 0$ then we have the equation $1/|w|^2 - \alpha /\bar w - \bar \alpha / w = 0$ . Thinking of $|w|^2 = w \bar w$ and multiplying out we get $1 - \alpha w - \bar \alpha \bar w = 0$ . Since $\alpha w + \bar \alpha \bar w = 2\Re (\alpha w)$ it means $\Re (\alpha w) = 1/2$ . If we let $\alpha = a+bi$ and $w=x+iy$ then it means we must have $ax - by = 1/2$ . And this is of course an equation of a line (since $a,b$ are not both zero because we are assuming $|\alpha| = r$ ).

Say that $r^2 - |\alpha|^2 \not = 0$ then we have the equation $1/|w|^2 - \alpha/ \bar w - \bar \alpha / w = r^2 - |\alpha|^2$ . Multiplying out and dividing (since it is non-zero) we get, $w\bar w - \beta \bar w - \bar \beta w = -(|\alpha|^2 - r^2)^{-1}$ where $\beta = \bar \alpha ( |\alpha|^2 - r^2)^{-1}$ . Add $|\beta|^2$ to both sides to get, $w\bar w - \beta \bar w - \bar \beta w + |\beta |^2 = r^2(|\alpha|^2 - r^2)^{-2}$ . Thus, $|w - \beta|^2 = r^2(|\alpha|^2 - r^2)^{-2}$ and this is an equation of a circle.
We proven that circles get mapped to lines or circles (depending whether a circle passes through origin or does not). Now we will prove lines get mapped to lines or circles. Let $S$ be a set which is a line. Then there exists $a,b,c$ (with $a,b$ not both zero) so that if $z=x+iy\in S$ then $ax+by=c$ . If we let $\alpha = a - bi$ then $ax+by=c$ implies $\alpha z + \bar \alpha \bar z = 2c$ and so $z=1/w$ which leads to $\alpha / w + \bar \alpha / \bar w = 2c$ thus $\alpha \bar w + \bar \alpha w = 2c |w|^2$ . This is either a circle or a line i.e. depending whether $c$ is zero or non-zero i.e. whether the line $ax+by=c$ passes through the origin or does not. This completes the proof.

There are two points here. First, given any Mobius transformation we can determine easily whether it will map a line into a line or circle. If for example $f(z) = z/(z+1)$ then any line which contains the point $-1+0i$ will be mapped into a line because the closer we get to this point the denominator of $f(z)$ becomes unbounded, that has to be a line because lines are unbounded. While if $-1+0i$ is not contained on the line then $f(z)$ does not have a vanishing denominator which will make the mapped curved bounded, so that is a circle. Similarly, if we are given a circle using that strategy we can determine whether it will map the circle into a circle or a line. Second, again I was not careful about the possibility about the circle passing through the origin in the proof above which will make, so it does not really map a line into a circle but that is nothing serious as was explained in the previous posts.

Seems almost too simple! What is the theorem called ? and how is it proved? If the proof is very detailed don't bother posting it,

The theorem does not have any name to it, it is a known result from complex analysis. The proof happens to be simple. Remember a curve in $\mathbb{C}$ is expressible as $z(t) = x(t)+iy(t)$ where $a\leq t\leq b$ (here $a,b$ are endpoints of the curve). Thus, $\cos t + i\sin t$ and $0\leq t\leq \pi$ is a semi-circle curve. We define $\dot z (t) = x'(t) + iy'(t)$ (if $x(t),y(t)$ are differenciable). You may wonder why not write $z'(t)$ instead. That is because the derivative of a complex function and the derivative of a curve are similar concepts but they are defined a little differently and so avoid a possible misunderstanding I find it better to use a dot rather than a prime to denote a curve derivative rather than a function derivative (this might be uncommon in some places so if you see other people do it differently do not get confused). Let $f(z)$ be an analytic function having a non-vanishing derivative at $z_0$ if $z(t)$ is a curve $C$ so that $z(t_0) = z_0$ then $F(t) = f(z(t))$ is $f(C)$ . Note $\dot F(t_0) = f'(z(t_0))\dot z (t_0) = f'(z_0)\dot z(t_0)$ . Thus, $\arg \dot F (t_0) = \arg f'(z_0) + \arg \dot z(t_0)$ . This completes the proof because this is shows that the argument of the tangent line (the argument between the tangents is the angle between the curves) after mapping it from $C$ to $f(C)$ gets increased by $\arg f'(z_0)$ , i.e. a constant number independent of the curve. Therefore, if two curves intersect at $z_0$ then applying $f(z)$ to each curve will increase the angles of both curves by the same amount, henceforth the angle between them does not change! Does that make sense?

You might be wondering why we need $f'(z_0)\not = 0$ . That is because the argument of $0$ is simply not defined, and so it does not make sense to talk about angles between curves.

I can always purchase a book on the topic, could you recommend a good one though ?

I would not recommend any book for you now. Complex analysis is much easier if you are familar with some basic analysis and advanced calculus. But if you really want to because you are impatient you can buy Complex Analysis by John Howie, that was the first one I read, it made as simple as possible but yet with enough mathematical rigor. The book I am using now (and the above proofs I wrote are based on it) is Complex Analysis by Bak and Newman (mathematicans are not very creative about naming books, so it is no wonder you can have 10 different books all called Complex Analysis). However, this book is more challenging than the first one I mentioned because it assumes the reader much sufficient analysis. Then there is a book called Visual Complex Analysis by Needham (or something like that), I never read the book (but plant to) so I cannot say. I heard it is an excellent book. Though the negative side with this book is that it is not a traditional book on complex analysis, it rather focuses more on the geometric side of complex analysis (which is what you are learning now). This means it is probably intended for students who already know complex analysis.

**ThePerfectHacker** · May 6th 2008, 10:52 AM

Mobius transformations can be visualized in a very nice way. The idea was introduced by Bernhard Riemann. Take a sphere of radius 1 and place it on the complex plane, directly on the origin. Look at the picture here. Let $N$ be the north pole, and $P$ be any point on the plane. Draw a line segment $NP$ , it shall intersect the sphere at a point $M$ . This point $M$ is the corresponding point to $P$ . And conversely, given any point on the sphere (not including the north pole!) the line joining the north pole with this point will intersect the plane at one point, this point will be the corresponding point. The link shows this in detail. Therefore, every single point in the complex plane can be visualized as every point on the Sphere (called Riemann Sphere), except of course the north pole.

The next thing we do is extend the complex plane. We define $\mathbb{C}_{\infty} = \mathbb{C}\cup \{ \infty \}$ . What is $\infty$ , you might ask? It is just a symbol. It represent the North Pole. The idea is that the closer a point moves to the north pole on the Riemann Sphere the larger its corresponding point becomes. Therefore we are motivated to use $\infty$ as the symbol for the North Pole.

We however do not define what $\infty + a$ is or what $a\cdot \infty$ is. We leave algebraic notions alone, i.e. we do not extend notions of addition, multiplication, subtraction, and division to $\mathbb{C}_{\infty}$ . Eventhough, we do not define any arithmetic (algebra) on $\mathbb{C}_{\infty}$ it is still a useful notion. When you will study topology you will learn that to do topology one does not need to have algebra on the set he is working with. The same idea is here, that eventhough we avoid algebraic issues we can still have a useful notion.

Suppose we have the function $f(z) = 1/z$ . This function is not defined everywhere on $\mathbb{C}$ . We can extend this function to $\mathbb{C}_{\infty}$ so that is will be defined everywhere. We do this by defining,
$f:\mathbb{C}_{\infty}\mapsto \mathbb{C}_{\infty}$ as $f(z) = \left\{ \begin{array}{c}\frac{1}{z} \mbox{ if }z\in \mathbb{C} - \{ 0 \} \\ \infty \mbox{ if }z=0 \\ 0\mbox{ if }z=\infty \end{array} \right.$ .

Note what we did. It is as if we defined $1/0=\infty$ . But remember, I said we are not defining any arithmetic for $\mathbb{C}_{\infty}$ . All we did was to redefine the function without ever raising the questions of arithmetic. Similarly, it is as if we defined $1/\infty = 0$ . But again if you look carefully we never defined any arithmetic we simply redefined a function. This happens to be a very nice idea, because it is as if we are defining what dividing by $0$ or $\infty$ is without every running into the problem of trying to define this type of arithmetic.

There is another nice feature. In topology there is a notion of continous functions in general, not just $\mathbb{R}$ or $\mathbb{C}$ . It turns out that the way we defined the function $1/z$ on $\mathbb{C}_{\infty}$ happens to be continous.

Here is another example, if we had $f(z) = (z+1)/(2z+1)$ we can extend this to,
$f:\mathbb{C}_{\infty} \mapsto \mathbb{C}_{\infty}$ defined as $f(z) = \left\{ \begin{array}{c} \frac{z+1}{2z+1} \mbox{ if }z\in \mathbb{C}-\{ \frac{1}{2} \} \\ \infty \mbox{ if }z=-1/2 \\ \frac{1}{2} \mbox{ if }z=\infty \end{array} \right.$ .

In fact every Mobius transformation can be extended in a similar manner to the Riemann Sphere.

The notion of the Riemann Sphere leads us to another nice feature. If $f,g$ are Mobius transformations on $\mathbb{C}$ we cannot necessarily compose $f\circ g$ at every point because the domains of $f$ and $g$ might not match up. However, if we think of $f,g$ as being extened to $\mathbb{C}_{\infty}$ then it is always fine to compose $f\circ g$ at every single point.

The above paragraph has some nice implications. You probably heard of "Group Theory" before. A group $G$ is a set with an operation $*$ so that: (i) $a*b\in G$ for any $a,b\in G$ (ii) $a*(b*c) = (a*b)*c$ for all $a,b,c\in G$ (iii) there is an "identity element" $e$ so that $a*e = e*a = a$ (iv) for any element $a\in G$ there is an "inverse" $b$ so that $a*b=b*a = e$ . For example, the integers $\mathbb{Z}$ are a group under the operation $+$ . Now if we let $\mathcal{M}$ be the set of all Mobius transformation on the Riemann Sphere then $\mathcal{M}$ is a group under $\circ$ (function composition). Note, we could have not made such a statement about a group if the Mobius transformations were defined only on $\mathbb{C}$ because as stated above we cannot necessarily compose them. If you want you can try proving that $\mathcal{M}$ , the "Mobius group", is actually a group.

Now we get to the geometry on the Riemann Sphere. Let a plane (not tangent to the sphere) pass through the sphere. Then the intersection would be a circle. So a "circle" on a Riemann Sphere is exactly a circle on the sphere. Here is where it gets interested. Let $l$ be a line in the complex plane. If you correspond all points of $l$ to all points of the Riemann sphere you will get a circle passing through $\infty$ . The idea being is that since the line is infinite in length it must pass through the north pole for the closer we reach the north pole the bigger we get. But there is more! Let $c$ be a circle in the complex plane. If we correspond the circle to the Riemann sphere we will get a circle not passing through $\infty$ .
The video below illustrate what I said. There is one minor detail. The above explanations are based on a sphere which sits on top of the plane. In this video the sphere's center coincides with the center of the plane. Some people find it more pleasing when the sphere and plane both pass through the center. But that is nothing serious. The same idea of correspondense applies.
[YOUTUBE]sk12ElESpQw[/YOUTUBE]

Finally we can easily describe all Mobius transformations as simple transformations of the Riemann Sphere. Let $P$ be a point in the complex plane, let $P'$ be its corresponding point on the Sphere. Rotating the sphere by $\theta$ will give a new point $P''$ . Correspond this point back to the plane, and get $P'''$ , it will turn out that $P'''$ is the point after rotating $P$ by $\theta$ . Therefore, rotations of the plane are simply rotations of the Sphere. Let $Q$ be a point in the complex plane, let $Q'$ be its corresponding point. If you raise the sphere 1 unit up, so that it floats above the plane (here I am thinking of Sphere as sitting on top of the Sphere like I been doing in the beginning). And now if you correspond (project) the raised point $Q'$ back onto the plane you will get a dilation of factor of $2$ . Therefore, dilations can be visualized as raising and lowering the sphere. Translations are the easiet one. Move the sphere around this will move the corresponding point arround. But it is the inversion which is by far the most interesting transformation. It can be proven (maybe you want to try to show this) that if $(\alpha,\beta,\gamma)$ is a coordinate on the sphere (so $\alpha^2+\beta^2 + (\gamma - 1)^2 = 1$ ) then $(\alpha,\beta,1-\gamma)$ will be the corresponding point after the mapping $1/z$ . Therefore, all the points on top go below the Sphere and all the points below the Sphere go above the Sphere. More precisely, it means we take the Sphere and flip it over, i.e. we "invert" it (hence the name "inversion").
Here is probably the best video I seen on YouTube:
[YOUTUBE]JX3VmDgiFnY[/YOUTUBE]

The video is illustrating everything what I said. It also is showing that since every Mobius transformation is a composition of the basic transformations we can think of any Mobius transformation as simply moving the Sphere around.

We can now prove the theorem that $1/z$ map lines and circles into lines and circles in a very geometric way.
We have four cases, for a circle on the Sphere. Remember the south pole is the origin of plane.
1)Passes through the north and south pole.
2)Passes through the north but not the south.
3)Passes through the south but not the north.
4)Passes neither through the south nor north.

In #1, we have a line passing through origin then $1/z$ is flipping the sphere over, which will still be a line through the origin because the north becomes the south and the south becomes the north.

In #2, we have a line not passing through the origin. Then $1/z$ will take this line to a circle.

In #3, we have a circle through the origin. Then $1/z$ will take this circle to a line because the new circle will pass through the north pole, and lines as circles which pass through the north pole.

In #4, we have a circle not passing through the origin. Then $1/z$ will take this circle on the Sphere and still keep it a circle on the Sphere, not passing through the origin. In this case we have another circle.

**bobak** · May 6th 2008, 04:31 PM

Originally Posted by ThePerfectHacker

The above paragraph has some nice implications. You probably heard of "Group Theory" before. A group $G$ is a set with an operation $*$ so that: (i) $a*b\in G$ for any $a,b\in G$ (ii) $a*(b*c) = (a*b)*c$ for all $a,b,c\in G$ (iii) there is an "identity element" $e$ so that $a*e = e*a = a$ (iv) for any element $a\in G$ there is an "inverse" $b$ so that $a*b=b*a = e$ . For example, the integers $\mathbb{Z}$ are a group under the operation $+$ . Now if we let $\mathcal{M}$ be the set of all Mobius transformation on the Riemann Sphere then $\mathcal{M}$ is a group under $\circ$ (function composition). Note, we could have not made such a statement about a group if the Mobius transformations were defined only on $\mathbb{C}$ because as stated above we cannot necessarily compose them. If you want you can try proving that $\mathcal{M}$ , the "Mobius group", is actually a group.

firsly we define the Mobius set as the set of functions $\forall a,b,c,d \in \mathbb{C}$ with $ad-bc \neq 0$ such that $f(z) = \left\{ \begin{array}{c} \frac{az+b}{cz+d} \mbox{ if }z\in \mathbb{C} - \{ - \frac{d}{c} \} \\ \infty \mbox{ if }z=- \frac{d}{c} \\ \frac{a}{c} \mbox{ if }z=\infty \end{array} \right. $

for i) I defined $f(z) = \left\{ \begin{array}{c} \frac{az+b}{cz+d} \mbox{ if }z\in \mathbb{C} - \{ - \frac{d}{c} \} \\ \infty \mbox{ if }z=- \frac{d}{c} \\ \frac{a}{c} \mbox{ if }z=\infty \end{array} \right. $

and $g(z) = \left\{ \begin{array}{c} \frac{ez+f}{gz+h} \mbox{ if }z\in \mathbb{C} - \{ - \frac{g}{h} \} \\ \infty \mbox{ if }z=- \frac{g}{h} \\ \frac{e}{g} \mbox{ if }z=\infty \end{array} \right. $

$f \circ g = \frac{(ae+bg)z+af+bh}{(ce+dg)z+hd+cf} $ which is an memeber of the set $\mathcal{M} $

ii) you require the composition function to a associative, before diving into messy algebra I noticed something about the first part.

if I write the coefficient of $f$ a matrix, and the coefficient of $g$ in a matrix, the coefficients of $f \circ g$ is $\left(\begin{array}{cc}a & b \\c & d\end{array}\right) \times \left(\begin{array}{cc}e & f \\g & h\end{array}\right)$

so is sufficient to state that matrix multiplication is associative so therefore the composition function is?

iii) the identity element is $f(z) = z$

iv) and the inverse element is $f^{-1}(z) = \left\{ \begin{array}{c} \frac{b+dz}{cz-a} \mbox{ if }z\in \mathbb{C} - \{ - \frac{d}{c} \} \\ - \frac{d}{c} \mbox{ if }z=\infty \\ \infty\mbox{ if} z=\frac{a}{c}\end{array} \right. $

This any good?

I will hopefully prove that thing about coordinates of the sphere after the inverse transformations tomorrow.

Thanks TPH

Bobak

**ThePerfectHacker** · May 6th 2008, 06:10 PM

Originally Posted by bobak

This any good?

Almost.

There is a minor problem here:

$f(z) = \left\{ \begin{array}{c} \frac{az+b}{cz+d} \mbox{ if }z\in \mathbb{C} - \{ - \frac{d}{c} \} \\ \infty \mbox{ if }z=- \frac{d}{c} \\ \frac{a}{c} \mbox{ if }z=\infty \end{array} \right. $

If you want to define it carefully there two cases:
Case 1)If $c\not = 0$ then $f(z)$ is exactly how you defined it.
Case 2)If $c=0$ then it must mean $d\not =0$ (the determinant is non-zero). Thus, $f(z) = \left\{ \begin{array}{c} \frac{az+b}{d} \mbox{ if }z\in \mathbb{C} \\ \infty \mbox{ if }z=\infty \end{array} \right.$ .

Also when you got $f^{-1}(z)$ you need to show that it belongs to $\mathcal{M}$ . But that is real easy.

And yes Mobius transformations can be regarded are matrix multiplcations. In fact, that is how I remember the formula for $f^{-1}$ .

**ThePerfectHacker** · May 7th 2008, 03:49 PM

Now we can get to something more interesting. And that is conformal mapping. So far we gave examples of conformal maps. Before going into the mathematics first we should understand what conformal maps are about. There are such functions that are called "isometric maps" and those are the functions (on $\mathbb{C}$ ) which leave shapes intact. For example, rotations, reflections, translations. More precisely a function $f$ is "isometric" if and only if $|z_1-z_2|= |f(z_1)-f(z_2)|$ . This is a fancy way of saying is that if you have two points $z_1,z_2$ then the distance of those two point is unchanged under $f$ . Since $|z_1-z_2|$ measures the distance of two points the statement $|z_1-z_2|=|f(z_1)-f(z_2)|$ means exactly what we said. For example, $f(z) = e^{i\theta}z$ this is a rotation by angle $\theta$ . And of course rotations leave a shape intact. More precisely it means the distances of two points are preserved. To see this, $|f(z_1)-f(z_2)| = |e^{i\theta}z_1 - e^{i\theta}z_2| = |e^{i\theta}||z_1-z_2| = |z_1-z_2|$ because $|e^{i\theta}|=1$ . A natural question to ask, are there any more isometric maps except for translations, rotations, reflections. And the answer is very supprising. It is that there is just one more type of isometry and that is a glide reflection. Which is supprising because we have billions and billions types of functions and there are just 4 different types of functions which are isometries. Isometries of course are conformal because since they leave shapes intact they would certainly leave angles intact. If we are given two regions we might not necessarily find an isometry mapping one onto another. For example, let $\Omega_1 = \{ |z| < 1\}$ and $\Omega_2 = \{ \Re (z) > 0\}$ . There is just no way how an isometric map can map $\Omega_1$ onto $\Omega_2$ . But, eventhough we might not find a distance-preserving map (isometry) we can still find an angle-preserving map. So the point is eventhough we can map $\Omega_1$ onto $\Omega_2$ at least in such a way to that angles are preserved. Let us try another example, let $\Omega_3 = \{ 0<\Im (z) < \pi \}$ and $\Omega_4 = \{ \Im (z) > 0\}$ . As you can see $\Omega_3$ is an "infinite strip" while $\Omega_4$ is a "half-plane" so there is no way to find an isometry between them. But when we map one onto another we would like to preserve some property, can we do it with angles? The answer is, yes! I leave it to you to verify that $f(z) = e^z$ is the required map. Surly, there got to be some really strange blob regions which cannot be conformally mapped between themselves. For example, take your pen and scribble a crazy looking region. Of course we cannot map that isometrically onto the a unit disk, but can we map it conformally? But maybe we are asking too much. Meaning, maybe we cannot even find a function that is going to map one onto another. In that case then are conformal map cannot exist because there is not even a function sending one onto another! The answer is unbelievable. Bernhard Riemann in his PhD thesis in 1851 showed that not only can you find a function mapping one onto another you can find a function which will be conformal! This happens to be one of the most impressive theorems in analysis, called Riemann mapping theorem. At this point we can state the Poincare conjecture. The Poincare Conjecture (proven in 2006) states that if you have two regions in three-dimensional space, you can map one onto another. However, unlike the Riemann mapping theorem it does not gaurentee that such a mapping in three-dimensions will state conformal. Therefore, you can think of the Riemann mapping theorem as a stronger version of Poincare conjecture in two-dimensional space. Of couse the mapping theorem is one of those theorems in math which tell us that a function exists, how to actually find it, is a different story. Therefore, we will study certain special conformal maps, and see some examples how to map regions onto different regions. But before doing that you might ask why would we want to find a conformal map in the first place? What is so special? Conformal mappings have applications outside of math. For example, if we take the globe of the world. And suppose we want to unravel it and lay it flat on the table. Can we do it? Gauss' Theorema Egregium from Differencial Geometry tells us that no matter how smart we are we can in no way lay the globe flat onto the plane. More mathematically, it means we cannot find an isometric mapping from the globe onto the plane. If you realized all the known maps (here map as in the "world map") that are used are distorted near the top and bottom, see this. So the sad news is we cannot find a perfect world map. However, we can still find a conformal map (try not to get confused which "map" I am talking about

) so that even though the map gets slightly distorted it still has its angles preserved. And that is an advantage to anyone who is using the map. So that is one example of how conformal maps are used, quite simply, conformal maps are used for making maps (what a great pun). There are other uses. Such as brain mapping, but I do not know anything about that. Also, in hydrodynamics the fuild flow can be modeled with the help of conformal maps, again I do not know anything about that. So our upcoming goal is going to be to explore different types of maps, and try to construct maps between certain regions.

**ThePerfectHacker** · May 11th 2008, 09:02 PM

Here are some additional simple conformal maps.

1)Consider the function $f(z)=e^z$ . We can show that $f(z)$ maps $\Omega_1 = \{ 0 < \Im (z) < \pi \}$ onto $\Omega_2 = \{ \Im (z) > 0\}$ . To see this let $x+iy\in \Omega_1$ then $-\infty<x< \infty \mbox{ and } 0<y<\pi$ . Thus, $f(x+iy) = e^x e^{-y}$ . Note $0<e^x<\infty$ as $x$ ranges from $-\infty$ to $\infty$ . And $e^{iy}$ is an angle ranging from $0$ to $\pi$ . Thus, $e^x e^{-y}$ is the set $\{ r e^{i\theta} | r>0 \mbox{ and }0<\theta < \pi$ . This is precisely the upper half plane $\Omega_2$ .
Use this mapping whenever you want to map an "infinite strip" onto a "half-plane".

2)Consider the function $f(z) = z^2$ . Let $\Omega_1 = \{ \Im (z) > 0\}$ and $\Omega_2 = \mathbb{C} - [0,\infty)$ , i.e. "the cut along the non-negative axis". We can show that $f(z)$ will map $\Omega_1$ onto $\Omega_2$ . Let $re^{i\theta} \in \Omega_1$ , then $r>0$ and $0<\theta<\pi$ . Then, $\left( re^{i\theta} \right)^2 = r^2 e^{i\theta}$ . Note $r^2$ still ranges from $0$ to $\infty$ . But $2\theta$ ranges from $0 < 2\theta < 2\pi$ . The important point is, we never get the whole plane there is a cut. (It can be proven it is impossible to find a conformal map onto the whole complex plane, but that is beyond the scope of what we are doing). So basically $z^2$ magnifies the region-angle by factor of $2$ . A similar situation for $z^n$ .

3)We never defined $\log z$ . We will define for $z\not =0$ , $\log z = \ln |z| + i \arg (z)$ , where $\arg z$ is measured on the interval $(-\pi, \pi]$ . I leave it for you to verify that $f(z) = \log z$ maps $\{ \Im (z) > 0\}$ onto $\{ 0 < \Im (z) < \pi \}$ . Note, it is the reverse of what $e^z$ does.

4)We can define $z^{1/2} = e^{\log (z)/2} = |z|^{1/2} e^{i\arg (z)/2}$ . This function is similar as in #2. Unlike $z^2$ which maginifies angles of the region by two, this one shrinks them by two. Consider $\Omega_1 = \{ \Im (z) > 0\}$ , note if $re^{i\theta} \in \Omega_1$ then $r>0$ and $0 < \theta < \pi$ . This means $\sqrt{ re^{i\theta}} = \sqrt{r} e^{i\theta/2}$ by how we defined square roots. Note $\sqrt{r}>0$ and $0< \theta/2 < \pi /2$ . Thus, $\Omega_1$ gets mapped onto $\Omega_2 = \{ \Im (z) , \Re (z) > 0 \}$ , i.e. the "first quadrant". Note, $z^{1/2}$ shrunk the range $0 < \theta < \pi$ to $0 < \theta < \pi/2$ . This is what I meant it shrinks the angle of the region. Similarly $z^{1/3}$ will shrink stuff by $1/3$ .

Let us make a summary of what we know.

$kz$ is used for magnifing at region (dilation)
$z+\alpha$ is used for translating a region
$e^{i\theta}$ is used for rotating a region
$(1-z)/(1+z)$ is used for mapping the disk to a half-plane
$(1-z)/(1+z)$ is used for mapping a half-plane to the disk
$e^z$ is used for mapping an infinite-strip to a half-plane
$\log z$ is used for mapping a half-plane onto an infinite-strip
$z^n$ is used for increasing the angle of the region by $n$ times
$z^{1/n}$ is used for decreasing the angle of a region by $n$ times

Furthermore, all these functions have non-vanishing derivatives, which means they are conformal.

Exercises.
1)Find $\log (i)$ .

2)Find $\sqrt{-1}$ and $\sqrt{i}$ .

3)Find a conformal map from the unit disk to the "infinite triangle" $\{ re^{i\theta} | r>0 , -\pi/4 <\theta < \pi/4\}$ .

4)This one is harder. Find a conformal map from the unit disk minus the cut $[0,1)$ (i.e. Pacman with his mouth closed) to the entire unit disk. It might seem impossible, but remember what Riemann mapping theorem tells us.

**bobak** · May 20th 2008, 02:21 PM

Originally Posted by ThePerfectHacker

Exercises.
1)Find $\log (i)$ .

2)Find $\sqrt{-1}$ and $\sqrt{i}$ .

3)Find a conformal map from the unit disk to the "infinite triangle" $\{ re^{i\theta} | r>0 , -\pi/4 <\theta < \pi/4\}$ .

4)This one is harder. Find a conformal map from the unit disk minus the cut $[0,1)$ (i.e. Pacman with his mouth closed) to the entire unit disk. It might seem impossible, but remember what Riemann mapping theorem tells us.

1) very simple using the definition $\log (i) = i \frac{\pi}{2}$

2) using the definition of the function again $\sqrt{-1} = i$ and $\sqrt{i} = e^{i \frac{\pi}{4}}$

3) This is a composite, First map the unit disc to the right half of the plane and then shrink the angle. so we have

$f_{1}(z) = \frac{1-z}{1+z}$
$f_{2}(z) = \sqrt{z}$

so the mapping is $f_2 \circ f_1 = \sqrt{\frac{1-z}{1+z}}$

4) This is is a lot harder than the rest, I spend a while trying to think of something decent. I don't really understand the Riemann mapping theorem but I appreciate what you said about it and therefore if it is true then a comformal function exists. my best guess is $f(z) = z^k \ \ k>1$ but that is probably nonsense.

Bobak

**ThePerfectHacker** · May 20th 2008, 04:32 PM

Originally Posted by bobak

3) This is a composite, First map the unit disc to the right half of the plane and then shrink the angle. so we have

$f_{1}(z) = \frac{1-z}{1+z}$
$f_{2}(z) = \sqrt{z}$

so the mapping is $f_2 \circ f_1 = \sqrt{\frac{1-z}{1+z}}$

4) This is is a lot harder than the rest, I spend a while trying to think of something decent. I don't really understand the Riemann mapping theorem but I appreciate what you said about it and therefore if it is true then a comformal function exists. my best guess is $f(z) = z^k \ \ k>1$ but that is probably nonsense.

The Riemann mapping theorem says you can find such a conformal mapping. It is telling you it is possible, it does not actually say how to do it. Yes, this is a harder problem. Try some more problems, then return back to that problem.

1)Map the strip $\{ -2<\Re (z) < 1\}$ to the unit disk.

2)Map $\{ \Re (z) < \pi/2 , \Im (z)>0\}$ to $\{ x+iy : y<x, y>0\}$ .

Hint for the hard problem: Map Pacman to the upper unit disk first.

**bobak** · May 21st 2008, 03:33 AM

Originally Posted by ThePerfectHacker

1)Map the strip $\{ -2<\Re (z) < 1\}$ to the unit disk.

$f_{1}(z) = z + 1/2$ to translate first to get the strip $\{ -3/2<\Re (z) < 3/2 \}$
$f_{2}(z) = iz$ to rotate the strip giving $\{ -3/2<\Im (z) < 3/2\}$
$f_{3}(z) = \frac{\pi}{3} z$ to dilate the strip onto $\{ - \pi /2<\Im (z) < \pi /2 \}$
$f_{4}(z) = e^z$ to map this onto $\Re (z) > 0$
$f_{5}(z) = \frac{1-z}{1+z}$ to map the half plane onto the unit disc.

we want $f_5 \circ f_4 \circ f_3 \circ f_2 \circ f_1 = \frac{1 - e^{i \pi/3 (z + 1/2)}}{1 + e^{i \pi/3 (z + 1/2)}}$

Originally Posted by ThePerfectHacker

2)Map $\{ \Re (z) < \pi/2 , \Im (z)>0\}$ to $\{ x+iy : y<x, y>0\}$ .

$f_{1}(z) = z -\pi /2$ to shift the region into the second quadrant.
$f_{2}(z) = -iz$ to rotate the region to the first quadrant.
$f_{3}(z) = \sqrt{z}$ to shrink the angle of the region form $0 < \theta < \pi/2$ to $0 < \theta < \pi/4$