# Conformal Mapping Explained | Möbius Transformation | Complex Analysis #12

– [Teacher] Hello and welcome. In this video, we are going to talk about conformal mappings, which is

a concept in complex analysis, that I know can be really hard to wrap your head around at first glance. So that is why I am going

to try in this video to really break down all the theory behind conformal mappings, and

also show you step by step how to solve these kinds of problems by using different methods,

so that you can decide for yourself which one

that works best for you. And remember that you can

always find the timestamps and additional resources

in the description. So, without further ado, let’s get going. The first thing you need to know is that a conformal map

is just another name for a function which

fulfills some conditions. We say that a function is conformal if it preserves angles locally. So, in short, a conformal map is just a function that preserves angles. But this leads us to our next question, which is what angles are we

really talking about here? And I thought that I

would answer this question by presenting a slightly different version of this definition by drawing of it. So, here we have two

directed smooth curves, which intersects at the points Z naught. The angle between the

two curves at Z naught is not defined as one might think, as simply the angle

between the two curves. But instead, it is defined as the angle between the two tangent vectors

of the curves at Z naught. And note here that the tangent vectors are pointing in the directions consistent with the

orientation of the curves. Let’s continue by looking

at the image of the curves under sub-function F. The points in naught might have ended up around here somewhere, with the curves curving in something like this. And if we now, once

again, define the angle between these two new curves

at this new intersection with the help of a tangent vectors, we get the following image. And with that done, I can

finally state the following. A function F is conformal

at the point Z naught if the angles alpha and

alpha prime are equal for every pair of directed smooth curves that intersects at this point Z naught. This is in fact just

another way to illustrate the properties of a conformal mapping, which we stated in the first definition. But I hope that we here

can more clearly visualize the angles in question, the angles that need to be the same for the mapping to be conformal. Now we will continue with another theorem, where we once again will say that analytic functions are superior

to all other functions. Because this theorem states that an analytic function is always conformal at the point Z naught, if the derivative at this

point is not equal to zero. I will add a link to prove this

theorem in the description, but just take my word

for it in the meantime, because we will need this

information when we now continue to talk about Möbius transformations, where a Möbius

transformation is defined as any function that can be

written on this form here, where the coefficients A, B, C, and D, are complex numbers,

satisfying that A times D minus B times C is not equal to zero. If we look at a derivative

of the function, we can see why this

condition is important. Since here, we can see that if A times D minus B

times C is equal to zero, then the derivative will

also be equal to zero, which means that the function

above is on a constant, and will therefore map all

input values to the same point. But even more important

is that here we can see that the derivative of

a Möbius transformation is never equal to zero,

thanks to this condition. Therefore, according to our last theorem, we know that a Möbius

transformation is conformal at every point except at its pole, since this is the only

point where a function is no longer analytic. And this is one of the reasons

why Möbius transformations are normally used when solving problems about conformal mappings. And the next thing we

are going to talk about is how this function is connected with some of the most

elementary transformations. The first transformation

we are going to talk about is called translation, which

is when a function moves every point of a figure

by the same distance in a given direction. For example, in this image, we can see that this rectangle,

determined by the origin and the letters A, B, and

C, under this mapping, has been moved by the same

amount and in the same direction determined by the complex number B. Rotation is the next transformation we are going to talk about, and it is simply the act of rotating every point of a figure around the origin. The function that does this can be defined as the following. And here we can see that the

amount we are rotating by is determined by the angle alpha. The third transformation

is called magnification. And this transformation

simply enlarges or contracts the distance of every

point from the origin by the same factor. For example, this function here enlarges the figure by the factor two. But note also here that the

distance between any two points is multiplied by the same factor. The last transformation

is called inversion, and it is defined as the

function one divided by Z. And to understand what this function do, let’s continue by rewriting Z

as the following expression. And here you can visualize

the absent value of Z as the length of the complex number, while the second factor

corresponds to the orientation of the same complex number. If we now calculate one divided by Z we get the following. And here we can see that this mapping changed both the length

and the orientation of every point. The new length is going to be equal to one divided by the old length. And the new orientation

of the complex number is simply the old orientation, but with a change of sine. The good thing to know

about this transformation is that it always maps generalized circles to generalized circles, which

is just a fancy way of saying that it always maps a line or a circle to either a line or a circle. For example, by using this transformation, this line that passes through the origin will be mapped to another line, but this time with another orientation. Meanwhile, this circle that

does not touch the origin will be mapped to another circle. And note here that the key difference between these two cases is not that one of them is a circle, and the other one is a line. Instead, the key difference

is that one of them passes through the origin,

and the other one does not. Because inversion of circles and lines works the way that if the

original line or circle passes through the origin, then its image will contain the point at infinity, since we get one divided

by zero at this point. So the image must therefore be a line. But if the original line or circle misses the origin, then

its image will be bounded, hence, it must be a circle. These four transformations relates to the Möbius transformation by the fact that any Möbius transformation

can be expressed as the composition of a finite sequence of translations, rotations,

magnifications, and inversions. So this theorem tells us

that a Möbius transformation is only a combination of these

other four transformations. Which means that you

only need to understand these four different

kinds of transformations to be able to determine

a Möbius transformation. And I thought I would

continue by showing you exactly what I mean by that, by doing some examples. In this example, I would

like us to determine the Möbius transformation

that maps the rectangle with the following four

corners in the Z plane onto another rectangle with

these corners in the W plane. Note that the mapping must be done so that the point Z1 is mapped to W1, and Z2 to W2, and so on. Let’s start by drawing the two rectangles and mark all the corners. And I would also recommend you to always draw the two figures in question when trying to solve a

conformal mapping problem, since this makes it

more easy for us to see how we have to transform our first figure to get to the second one, by using these

transformations to your right. We can, for example, start by noticing that the rectangle in the W plane is orientated in a different way compared to our rectangle in the Z plane. They differ by a rotation of 90 degrees. But we can fix this by

rotating our figure 90 degrees, or two pi anti-clockwise. So by using this formula for rotation, we get the following. Which we can simplify to this, since E raised to a power

of I times pi divided by two is equal to I. The next thing that we can do is to adjust the size of the figure, since we can see here

that our new rectangle is only one long and half high. Meanwhile, the other one

is two long and one high. So we need to use magnification to make our new rectangle twice as big. Which we can do by using

the following function. And now you might wonder

why we’re using F1, and not Z in this formula. And that is because we are

transforming our new rectangle, and not the one we started with. And our new rectangle was a result of our first transformation, F1. So that is why we’re using F1, and not Z. Now we can see that the figure

seems to match pretty well, but we still need to

move our figure a bit up, which we can do with translation. To determine the amount we

need to translate it with, we can, for example,

observe that the point Z1 is located at origin, at the moment. But we would like it to be one step up along the imaginary axis right, since that is where W1 is

located in our last figure. By using this formula for translation, with B is equal to I, we get the following. And this one should move our last figure, which was the result of the function F2, one step up, along the imaginary axis. And now we have actually succeeded to map our rectangle in the Z plane to this new rectangle in the W plane, so that the corners corresponds correctly. But the last thing we need to do is to go backwards and add all the results from all of these

transformations together, so that our final answer is only expressed as one function, and not

as a composition of three. So by inserting the

expressions of the function F1 and the function F2 into our function F3, we get the following. And this thing is our final answer. This is our Möbius transformation, since a Möbius transformation

is always made up of a sequence of these

other four transformations. And that is why I said before that you only really need to

know these four transformations to be able to solve

these kinds of problems. Let us do another example. But this time, I would

like us to determine a Möbius transformation

that maps this circle here onto this circle. For starters, by drawing them both, we can note that the circles are different in regards to a size. The first one has a radius of one, where the second one has a radius of two. But the positions of

the circles also differ, since one of them is centered around one, while the other one is centered

around 3I divided by two. Let’s continue by moving

the circle to the origin, which we can do by using translation, with B is equal to minus one, since we want to move

the center of the circle one step to the left, along the real axis. And the next thing we would like to do is to adjust the size of the

circle by a factor of two, by using the formula for magnification. And once again, remember

that we are transforming the last figure, and not

the one we started with. And that is why we are using

the function F1 and not Z. Now that the size is correct, we just need to move it so

it is centered correctly. Our figure is centered

around the origin, right? But we would like it to be centered around the point 3I divided by two, as the circle is in the last image. And we can do that by once

again using translation. But this time by adding 3I divided by two to our last figure. Now we once again go backwards and combine the results of

each of the transformations to get the final expression, which will be our Möbius transformation. But note that this is not the

only answer to this problem. There is actually an infinite amount of Möbius transformations

that work for this problem. So let me know in the comments

if you can find another one. In this problem, we are going to determine a Möbius transformation

that maps the inside of the unit circle onto

the right half-plane. The major difference between this problem and the other ones we are done, is that here, we are dealing with an area inside of a figure, instead

of only the figure itself. And we are therefore going to need to keep track of this area as we go along. But we are still going to focus on transforming the circle, since the area is contained

inside the circle. First off, we need to

transform the circle to a line, which we can do by using inversion. But here we must remember

that to get the line, we need to make sure that we are choosing our expression for the function smart, since we need our circle

to pass through a point which makes our function divide by zero, since this will make the

image contain infinity, and therefore must be a line. We cannot get this result by using one divided by zero as our transformation, since the circle never

passes through the origin. But we can use one of the

following expressions, since the circle passes through points that makes each of them

divide by zero at least once. Note here that all of them

are valid transformations to transform our circle to a line. But they are going to

have different properties regarding where we place the line and how they map the

area inside the circle. But just choose one of them, and then we can adjust

these properties afterwards with additional transformations. So let’s continue by using the

function here, for example. Then the next step is to determine exactly how it maps our

circle to our new line. And we can do that by

inserting some points which lies on the circle, and see where the function maps them. The function maps the

point one to infinity, which is great, since now we know that we are dealing with

a line, as we wanted. And the function maps I to

minus one divided by two, minus I divided by two. And it maps minus one to

minus one, divided by two. And lastly, it maps minus I

to minus one divided by two, plus I divided by two. And by marking these

points here on the graph, we get the following line. And that’s great,

because we have succeeded in creating our line as we wanted. But now the million dollar question is, where did the area inside

the circle get mapped to? This area can only be

mapped to either a left or the right side of this line. And one way to determine this is to take one point

that makes up the area, let’s say the origin, for example. And then we see where it went. Here we can see that this point got mapped to minus one, right? So this means that the

whole inside of the circle is going to be mapped to

the left side of our line. Another way to determine the same thing is to go along the circle and note which side the area is on with respect of orientation, since this property will not

change after transformation. So if we, for example,

go from the top here, from I to minus one, and then to minus I, then we can note that the

area is to the left of us, with respect of orientation

we are traveling. And if we now travel along the same points after they have been remapped, then we know that the area

will still be on our left side, which also corresponds with

the result we got before. The next thing we have to adjust is the position of our figure, since the line goes along

the image in our axis in the last image. And we can do this by doing a translation, by adding one divided by two

to our last figure, right? So it should look something like this. And the last thing we need

to fix is that the area needs to be on the right

side of the line, right? And since we know that the area is directly connected

with the orientation, we can get this result

by flipping the line 180 degrees, or pi. Because now we will have

to go down instead of up to follow the same path as we did before. Which means that the

left side of the line, with respect to orientation

we are traveling downwards, which contains the area in question, will be on the right side

of the complex plane, which is exactly what we wanted. The Möbius transformation,

which we wanted to determine, can therefore be determined by going backwards, once again. So if we insert the expressions of the functions F2 and

F1 into our function F3, we get the following. And if we now write the two

divisions on the same line, we can see that we can

simplify the expression as the following. And here we can simply factor out one divided by two, which

will give us the following. And the final thing I would like to do is to factor out minus one in the numerator and denominator, since these two factors

eliminate each other. And now this thing here

will be our final answer. But now I’m going to

show you another method that you can use to solve

these kinds of problems. But note that this method can only be used if you know how three points are mapped, and this comes from the fact that a Möbius transformation

is uniquely determined by knowing two sets of the same points. The points Z1, Z2, Z3, whichever

point you are going to map, and the corresponding

points W1, W2, and W3, which are the same points, but after they have been mapped. The transformation is then given by solving this equation here, for W. And this equation comes from the fact that a Möbius transformation

always preserves the cross ratios. To the left side here, we have a cross ratio before the mapping. And to the right side, you have a cross ratio after a mapping for these points in question. And these two cross ratios are always going to be equal in the case of a Möbius transformation. Let’s continue by redoing

our first example, but this time by using

this method instead. So here we actually know how four different points are mapped. But since we only need three of them to use the formula, I’m going to discard the

pair Z4 and W4 in this case. If we now start by inserting

our starting points, which are the following Z

values, into our formula, we get the following. And if we now insert where

these points are going, which are the W values, into the formula, we get the following expression. And now we just have to

solve this equation for W, which is easier said than done. So, hang on tight. We can start by removing all the zeros in the left side of the equation. And we can also simplify the

right side of the equation by noticing that we

have I minus 2I up here, and down here we have I minus I. So if we simplify this,

we get the following. The next step is to get

everything on the same level, which we can accomplish by multiplying both sides of the equation with the other sides denominator, which will give us the following. If we now write out the multiplication, we get the following. And here we can see

that we have W times 2IZ and W times Z on both

sides of the equation. And we can therefore remove them, since they eliminate each other. So, we end up with a equation that looks something like this. And now we can factor out the W on the right side of the equation, which will give us the following. And now we can get the expression for W by dividing both sides with

I minus one divided by two. And we can simplify this expression by multiplying up and down here with the complex conjugate

of the denominator, since this will create a real

number in the denominator. So let’s see. In this case, we are going to end up with five divided by four down

here, as our real number. And by now expanding the numerator, we get the following. And here we can observe

that we have minus Z and Z. So they eliminate each other. But the same thing happens

with minus one divided by two, and one divided by two over here. And for my next step, I’m

going to sidetrack a bit, because here we need to note

that 2IZ plus IZ divided by two is equal to five divided

by four times 2IZ. And we also need to know

that I plus I divided by four is equal to five divided by four times I. And this is important, because

now we can use this knowledge to factor out five divided by four in both the numerator and the denominator, which will give us the following. And now, by doing one last simplification and removing the five

divided by four here, we get our final answer. Which we can note is the same answer as the one we got here above. So, there you have it. Two different ways of

solving the same problem. And it’s all up to you to

choose which one to use. This video became rather long, but I really hope that

you learned something new about conformal mappings, and that you feel like you

have some methods to use to tackle these kinds of problems. And just let me know in the comments if you have any questions. Thanks for watching. (mellow music)

Long-time no see, thank you all for the supportive messages during my hiatus (I completed my master thesis and have worked one year at my first job)! This video is dedicated to all of you, but also specific @Sarah _ that have waited an awful amount of time for me to complete this one.

Conformal mapping is one of the hardest concepts to grasp in complex analysis and I also struggled with it a fair bit in school but I think this video will be able to show you that it is easier than one thought!

It looks really good! What software do you use to make these videos?