# 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)

## 2 thoughts on “Conformal Mapping Explained | Möbius Transformation | Complex Analysis #12”

1. TheMathCoach says:

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!

2. YOLOSWAG420 says:

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