Composing functions that are conformal

[d2j2003]

Homework Statement

let f and g be conformal analytic functions with the range of f a subset of the domain of g. Show that the composition g(f(z)) is also conformal.

Homework Equations

The Attempt at a Solution

Well I know that if the function is conformal it is 1-1, analytic, and its derivative is non zero. So I'm thinking that I would need to show that since f'(z)≠0 and g'(z)≠0 then (g(f(z))'≠0 and then show that since f and g are analytic, g(f(z)) is also analytic. Then the composition will be both analytic and its derivative will not be zero so it will be conformal. Am I working in the right direction?

 

[Office_Shredder]

That's one way to do it. The other way is to just go straight with the definition and show that g(f(z)) preserves angles

 

[d2j2003]

Is it possible to show, in general, that g(f(z)) preserves angles? I thought we would have to have actual functions defined for that..

 

[d2j2003]

How would I prove the derivative part of this? I know that (g(f(z)))'=g'(f(z))*f'(z) and obviously f' is not zero but I'm not sure how to show with certainty that g'(f(z)) is not zero...

 

[Dick]

How would I prove the derivative part of this? I know that (g(f(z)))'=g'(f(z))*f'(z) and obviously f' is not zero but I'm not sure how to show with certainty that g'(f(z)) is not zero...

You are given that "the range of f a subset of the domain of g".

 

[HallsofIvy]

Is it possible to show, in general, that g(f(z)) preserves angles? I thought we would have to have actual functions defined for that..

Let P, Q, and R be three points such that angle PQR has measure $\theta$. Then, since f is conformal, P'=f(P), Q'=f(Q), and R'= f(R) are three points such that angle P'Q'R' has measure $\theta$. And, since g is conformal, P''= g(P'), Q''= g(Q'), and R''= g(R') are three points such that ...

 

[d2j2003]

Let P, Q, and R be three points such that angle PQR has measure $\theta$. Then, since f is conformal, P'=f(P), Q'=f(Q), and R'= f(R) are three points such that angle P'Q'R' has measure $\theta$. And, since g is conformal, P''= g(P'), Q''= g(Q'), and R''= g(R') are three points such that ...

Such that the angle P''Q''R'' is θ meaning that the composition is angle preserving. right?

 

[d2j2003]

You are given that "the range of f a subset of the domain of g".

can we say that since the range f(z) is in the domain of g then g(f(z)) is basically g(z)? and then it is non zero from there?

 

[Dick]

Let P, Q, and R be three points such that angle PQR has measure $\theta$. Then, since f is conformal, P'=f(P), Q'=f(Q), and R'= f(R) are three points such that angle P'Q'R' has measure $\theta$. And, since g is conformal, P''= g(P'), Q''= g(Q'), and R''= g(R') are three points such that ...

f being conformal doesn't imply that. The 'angle preserving' in conformal refers to the angles between intersecting curves. Not to angles between points. That only works if f maps lines to lines.

 

[Dick]

can we say that since the range f(z) is in the domain of g then g(f(z)) is basically g(z)? and then it is non zero from there?

g' is nonzero on its domain. f(z) is in that domain. I wouldn't say 'g(f(z)) is basically g(z)', that's not very true.

 

[d2j2003]

well when we say g(z), that z includes f(z) since the range of f is a subset of the domain of g... ie. g(z) takes values from its domain, which includes f(z).. right?

 

[Dick]

well when we say g(z), that z includes f(z) since the range of f is a subset of the domain of g... ie. g(z) takes values from its domain, which includes f(z).. right?

Right. Just say f(z) is in the domain of g. Don't say g(f(z)) is the same as g(z).

 

[d2j2003]

ok, got it. Thanks for your help