# Complex analysis

1. Mar 8, 2012

### arthurhenry

Two questions:
1)Quote comes from a textbook:

Each non-constant function analythic function with f(0)=0 is,in a small nbhd of 0, the composition of a conformal map with the nth-power map...The proof is given and I think I am comfortable with it..

My question is a lot simpler (I think): Can we say the same for a function such that f(a)=b?
I am kindly asking someone to explain why one loses no generality if we
assume a=b=f(a)=0.

2) If an analytic function is not zero in a nbhd, what can I say about the derivative there? i.e. what restriction does this impose on f' (f prime)

Around a circle that lies in the nbhd above we should have
# Zeros=Integral (f'/f)=0

Should this not say f'=0 on that circle?
Thank you

2. Mar 8, 2012

### morphism

1) If f(a)=b, then f(z) looks like b+(z-a)^n near a.

2) Could you explain why you think f'=0 on the circle?

3. Mar 8, 2012

### arthurhenry

I was thinking: if f' were to be never zero, then how can Integral[f'/f] around a circle be zero.
The integral should be zero as f has no zeros inside the circle...Is what I am thinking.

Than you for your time again

4. Mar 8, 2012

### morphism

The contour integral of a nonzero function can very well be zero. Take for instance f(z)=z-2 and integrate f'/f=1/(z-2) on the unit circle.

5. Mar 8, 2012

### arthurhenry

Somewhat embarrassing...

Thank you very much for the help

6. Mar 13, 2012

### Bacle2

Actually, if f is analytic and 1-1, then f' is not zero; if f'(z) is 0 , then factor:

f(z)-zo=z^k(g(z)) (i.e., use Taylor series, where k is the index of the zero of f ),

and g(z) is analytic and non-zero in a 'hood U of zo (the zeros of an analytic non-constant

function are isolated) . The idea is that these conditions on f allow you to define a local k-th root function

in the ball, and k-th roots are k-to-1, and so in particular not 1-1.

Under these conditions , take a

small ball B(z,r) around f(z). This allows you to define a log locally, so that you can

then define a k-th root of g(z) (technically, the ball is simply-connected, and does not

wind around 0). Then you can define a k-th root using the log, and k-th roots are not

1-1; they are actually 1-1. Maybe needs work, but it is late.