# Complex conjugate as a Mobius transformation

1. Jan 24, 2013

### iLoveTopology

Hi guys,

I am having a very stupid problem. I can't figure out what Mobius transformation represents T(z)=z*, where z* is the complex conjugate of z.

In my book we are learning about Mobius transformations and how they represent the group of automorphisms of the extended complex plane (Ʃ). [ NOTE: My book lists 4 generators for the automorphisms of Ʃ: 1) rotations R(z)=cz (c a complex number) 2) Inversions J(z)=1/z, 3) scaling S(z)=rz, (r a real number), and 4) translations T(z)=z+t. Complex conjugation T(z)=z* is NOT listed as one of the generators. I mention this because reading some other documents I see some people do list this as a separate generator. So I'm not sure if this is relevant]

In my book "anti - automorphisms" is brought up and they have the form:

T(z) = (az*+b)(cz* + d) where z* is the complex conjugate of z.

In the book they say, each anti-automorphism T is the composition of complex conjugation with an automorphism of Ʃ. Then they say - "complex conjugation being given by reflection in the plane through ℝ U {∞}). Geometrically I understand - but algebraically I don't. What is the Mobius transformation that will take a point z to it's complex conjugate z*?

If I try T(z)=1/z (the Mobius transformation where a=0, b=1, c=1, d=0), I don't see how this works - because 1/z = z*/|z|2, not just z*. How can I get rid of the |z|2? I don't see how I can because it's dependent upon my z and it's squared and the Mobius transformations are rational.

I feel completely stupid for not seeing this!

tl;dr

What is the Mobius transformation for T(z)=z* ? Again, geometrically I understand WHY there is one - because we're just doing a simple reflection of a point so this is just an automorphism of Ʃ and therefore since all automorphisms of Ʃ are represented by Mobius transformations there should be a Mobius transformation to represent this transformation. But I don't know what it is! Once I have T(z)=z*, then I can compose this with my "goal" R(z)=(az+b)(cz+d) to get (az*+b)/(cz*+d) but I'm just not sure how to get that initial T(z)=z*.

Last edited: Jan 24, 2013
2. Jan 26, 2013

### iLoveTopology

just in case anyone ever sees this, I think* I figured out why z* is listed as a Mobius transformation. Because think of this, Mobius transformations are mappings T(z):C* --> C* (where C* is the extended complex plane). We know that the extended complex plane is isomorphic to the Reimann sphere, where the distance of any point on the Reimann sphere to the origin is exactly 1. So say we have some complex number z. z = a+bi. Then it's complex conjugate, is z* = a-bi. We can represent z* as |z|^2/z. Here's my thinking. z is analogous to some point on the Reimann sphere, and if z is a point on the Reimann sphere, it's modulus (distance from origin) is 1 so z* = 1/z.

I hope this thinking is in the right direction because it's the only thing I can figure out

3. Jan 28, 2013

### joeblow

Complex conjugation is not an analytic map. All Mobius transformations are. Thus, you will never find a Mobius transformation representing complex conjugation. The automorphisms of the unit disk involve taking the conjugate of one of the fixed points, not the variable z.

4. Jan 28, 2013

### iLoveTopology

Oh... if its taking the conjugate of a fixed point , that actually makes sense and clears up so much confusion. But that is not what my book says which is why I'm confused. Perhaps you can take a look at what the author is saying and you will understand what he means and where I'm going wrong.

Here's a link to the piece I'm speaking about I found it on google books

http://books.google.com/books?id=_U...=ANTI-AUTOMORPHISMS complex functions&f=false

In case it doesn't go to the right place, its p. 19 where he speaks of anti-automorphisms.

edit:
Oh wait... I feel so stupid. He was pointing these anti-automorphisms out as something distinct from the automorphisms of the Reimann sphere, which are Mobius transforms. These aren't automorphisms of the sphere, so there are no MT to represent them. The answer is right there in the definition!!!

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