19 - Möbius transformatios, part II

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welcome to the second part of our

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lecture series on Mobius transformations

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in the fourth week of our course

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analysis of a complex kind remember that

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we define Mobius transformations or

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fractional linear transformations as

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functions of the form a Z plus B divided

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by C Z plus D or ABCD or complex

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constants such that ad minus BC is non

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zero which is just guarantees that F is

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not constant these are called Mobius

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transformations here are some facts we

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showed last class a Mobius

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transformation Maps the extended complex

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plane which is C together with a point

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at infinity to the extended complex

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plane we were therefore allowed to plug

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in infinity for Z and we get f of

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Infinity's a over C as long as C is non

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zero or F of infinity is equal to

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infinity if C is equal to 0 we should

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the f is a conformal map from C hath to

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C hat it preserves angles it's 1 to 1

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and it's on 2 if C is equal to 0 I mean

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might just as well assume that D is

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equal to 1 and then F of C is of the

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form a Z plus B these are the conformal

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Maps from C to C they mapped infinity to

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itself and therefore are conformal maps

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from C to C

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no his transformations Maps circles and

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lines to circles or lines meaning every

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circle is mapped to a circle or a line

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every line is mapped to a circle or a

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line finally we showed that given three

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distinct complex number C 1 Z 2 and Z 3

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there exists a Mobius transformation

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that map's Z 1 to 0 Z 2 to 1 and Z 3 to

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infinity and we can actually write down

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this Mobius transformation and it is

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given by this formula right here today

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we want to construct some actual Mobius

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transformations before we go there let's

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prove a few more facts the first one is

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that the composition of two mobius

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transform

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it's again a Mobius transformation and

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the second one is that the inverse of a

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Mobius transformation is a Mobius

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transformation let me quickly show you

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how you would prove this fact

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suppose F and G are Mobius

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transformations F of Z is AZ plus B over

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C Z plus D and G of Z is say e Z plus F

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over G Z plus h and now suppose we

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wanted to look at F of G of Z and show

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that that again is a Mobius

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transformation so again has this form

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how would you do that well F of G of Z

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means instead of Z we're gonna write G

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of C so that's a times G of Z plus B

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over C times G of Z plus D and now we

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plug in ez plus F over G Z plus h for G

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of Z and at the same time we're gonna

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multiply top and bottom by GZ plus H to

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get rid of the fraction within the

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fraction and then left with a times ez

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plus F plus B times GZ plus h divided by

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C times ez plus F Plus D times GZ plus h

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and now if you multiply through and

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collect terms you see it's again off the

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form something times Z plus something so

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I'm just gonna do this for the top and

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not for the bottom so this we get a z

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term and it consists of a II plus B G

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and a second term that is without any

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Z's a F plus B H the same thing happens

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in the denominator and you see that this

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is again the form of a Mobius

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transformation similarly you can find

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that the inverse

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Mobius transformation is a Mobius

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transformation you simply write W as a Z

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plus B over C Z plus D and you solve

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this for Z now when you solve this for Z

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you will see that Z is a function of W

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expressed in the form of a Mobius

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transformation now we're able to show

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the following given three distinct

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points z1 z2 and z3 and three other

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distinct points w1 w2 and w3 there

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exists a unique Mobius transformation

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that maps Z 1 to W 1 Z 2 2 W 2 and Z 3 2

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W 3 and it's very easy to see that given

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the facts we already proved here's how

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you do this first of all we know that

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there's a Mobius transformation f that

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map's Z 1 to 0 Z 2 to 1 and Z 3 to

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infinity similarly there's a Mobius

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transformation F 2 that does that for

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the double use that map's W 1 to 0 W 2

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to 1 and W 3 to infinity so here's F 1

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and here is F 2 and now all we have to

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do is look at F 2 inverse composed with

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F 1 what's that going to do F 1 Maps Z 1

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to 0 F 2 inverse Maps 0 to W 1 and

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therefore this composition F 2 inverse

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composed with F 1 Maps C 1 to W 1

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similarly F 1 Maps Z 2 to 1 F 2 inverse

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Maps 1 to W 2 so the composition F 2

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inverse composed with F 1 maps z2 2 W 2

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similarly Z 3 is mapped to W 3

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let's look at an example suppose we

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wanted to find the Mobius transformation

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that maps 0 to negative 1 I to 0 and

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infinity to 1 so in other words this is

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our z1 this is our w1 this is Z 2 this

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is W 2 and this C 3 and W 3 I want to

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show you two different ways of getting

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there the first way is to use the proof

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of the previous fact that we just showed

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there for construct a mapping F 1 that

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map's the Z's to 0 1 and infinity and to

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construct a map F 2 that map's the W's

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to 0 1 and infinity so here's the

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mapping F 1 that Maps disease therefore

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0 I and infinity to 0 1 and infinity and

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I just rode out exactly the mapping that

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we had constructed before we have some

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funny terms in there off the form Z

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minus infinity and I minus infinity we

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need to interpret those correctly

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thankfully they both occurred and they

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therefore cancel each other out so the

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mapping is really Z over I and you can

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now check the mapping Z over I indeed

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maps 0 to 0 it Maps I to 1 because I

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over I equals 1 and it Maps infinity to

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infinity because infinity over I is

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still infinity F 2 is the map that takes

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the W's so minus 1 0 and 1 and maps

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those 2 0 1 infinity and again we just

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took the mapping that we constructed

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earlier and plugged in these values

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negative 1 0 and 1 and out came the map

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Z plus 1 over negative Z plus 1 but we

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need F 2 inverse because we need to look

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at F 2 inverse composed with F 1 that's

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going to be our final mapping to find F

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2 inverse we need to set F 2 of Z equal

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to W and solve that equation for Z

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to solve this equation for Z we multiply

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both sides of the equation by this

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denominator minus Z plus 1 and multiply

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3 this gives us W times minus C plus W

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is equal to Z plus 1 and now we collect

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terms that have a Z in them on the right

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in terms with our Z on the left on the

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right behalf of Z here and a minus W Z

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over on the other side which we're going

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to bring over to the right hand side

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therefore ending up with Z times 1 plus

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W and on the left we have W and this one

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that we're going to bring over here so W

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minus 1 we can divide by 1 plus W and

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find that Z is w minus 1 over W plus 1

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so we now have F 2 inverse of W being W

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minus 1 over W plus 1 and we can

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therefore look at F 2 inverse composed

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with F 1 so here are the two maps again

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that we constructed F 1 and F 2 inverse

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and therefore F 2 inverse composed with

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everyone is the desired map F 2 inverse

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off F 1 of Z can be obtained by plugging

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F 1 of Z in here for W so we find F 1 of

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Z minus 1 over F 1 FC T plus 1 F 1 of Z

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is the map C over I we can plug that in

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so C over I minus 1 divided by Z over I

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plus 1 we multiply top and bottom by I

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and end that with Z minus I over Z plus

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I that was a rather complicated way of

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finding this mapping it was the weight

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that followed our proof sometimes the

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mathematical proof is a very slick proof

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but does not give a good constructive

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way of finding a mapping there's a much

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easier way to find this mapping that

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doesn't follow the proof but is much

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easier so let's look at another approach

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we know that the mapping F that we're

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looking for is of the form a Z plus B

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over C Z plus D we wanted to map 0 to

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negative 1 I to 0 and infinity to 1

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since we want F of I

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we can assume that a is nonzero because

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if a was equal to zero the numerator

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would be fixed namely B and it'll be

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impossible to plug an AI and get zero

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therefore a must be nonzero and because

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we can multiply each of these numbers a

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b c dy at constant without changing the

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mapping we can assume that a is equal to

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1 therefore f is of the form Z plus B

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over C Z plus D I just plugged in 1 here

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for a now F of I is equal to 0 so if I

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plug an i for Z I need to get 0 the

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numerator then becomes I plus B I plus B

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must be 0 so B needs to equal to

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negative I therefore F is of the form Z

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minus I over cz plus T now let's use the

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fact that F of infinity is equal to 1

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when you plug an infinity here you get 1

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over C so 1 over C needs to equal to 1

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which means C needs to equal 1 and

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finally since F of 0 is negative 1 if

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you plug in 0 we get minus I over d that

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needs to be equal to negative 1 so T

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needs to be equal to I therefore f of C

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is of the form Z minus I over St plus I

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the same function we had found before

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but this was much much easier

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now it turns out there are basically

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only three types of Mobius

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transformations every Mobius

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transformation is the composition of

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maps of these types and the only types

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there are are the maps of the form Z

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maps to a times C which is a combination

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rotation and dilation if you write a its

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length times e to the I theta

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multiplying by a means rotating by theta

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and stretching or shrinking by a factor

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of absolute value of a the second type

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is a translation Z maps to Z plus B and

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the third type is an inversion Z maps to

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one over Z and those are the three types

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that make up every Mobius transformation

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let's see why that is true so folks F is

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a Mobius transformation suppose first

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that F maps infinity to infinity we saw

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that then F is of the type a Z plus B

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but AZ plus B is a composition of two

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types of Mobius transformations first of

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all Z gets mapped to a times C which is

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the rotation dilation type and then we

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add B to AC which is a translation so AZ

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plus B is indeed a composition of these

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two Mobius transformations the rotation

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and a translation

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next suppose that F F infinity is not

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equal to infinity then F is truly of the

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form a Z plus B over C D plus B where C

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is nonzero and we can divide all the

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constants by C therefore F is of the

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form a over C Z plus B over C divided by

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Z plus D over C in other words we can

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make the constant in front of Z equal to

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1 and with that constant being equal to

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1 we can simply assume that Z is equal

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to 1 so f of Z is really of the form a Z

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plus B over Z plus D with a 1 in front

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of the C but now let's do a little trick

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in the numerator we're gonna add an A

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times

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right here and then subtract an eight

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times D and we add the a times D by

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factoring out an a times D plus D this

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is the B rehab before and then we

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subtract 80 but by doing so we can split

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up this fraction right here into two

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fractions the first one is a times Z

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plus D divided by Z plus D so that's

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simply a and the second one is B minus a

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D divided by Z plus T looking at this

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carefully we see again that the moas

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transformation corresponds to

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composition the first artists were

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adding D to Z that's a translation next

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or flipping so we're using an inversion

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to get the Z plus D into the denominator

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next we're going to add in this

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numerator B minus ad that's a

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multiplication by B minus ad so that's

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the dilation and rotation kind and next

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finally we're going to add 8 to the

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whole expression which corresponds to a

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translation so again an arbitrary Mobius

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transformation is the composition of a

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translation and inversion a dilation and

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rotation and another translation we

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mentioned earlier the fact that Mobius

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transformations map circles and lines to

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circles and lines but we haven't proved

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that fat yet how would you prove this

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well we can now use this compositional

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result and we therefore need to only

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show that these three standard types map

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circles and lines to circles and lines

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because every Mobius transformation is a

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composition of these three standard

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types and if each of these three

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standard types does what we wanted to do

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then a composition will do the same

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thing while clearly dilations rotations

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and translations preserve circles and

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lines as circles and lines in fact they

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preserve circles and circles and lines

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and lines and so all that's left to show

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is that the inversion one over Z maps

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circles and lines to circles and lines

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we saw the main ideas on how to do that

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during the last lecture we looked at a

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bunch of examples of circles and lines

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and showed what their images were and

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all you'd have to do now is take an

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arbitrary circle instead of a specific

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circle and look at its image under the

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map now that we understand Mobius

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transformations much better we are ready

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to study the riemann mapping theorem

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will do so in the next lecture