# Show the following (complex) function is analytic

1. Apr 19, 2010

### synapsis

1. The problem statement, all variables and given/known data
Given that f is analytic in the open disc of radius one around zero, and that f(0)=0 and mod(f(z))<=1 for ever z in D(0;1).

Define g(z) = f'(0) if z=0, f(z)/z elsewhere.

I want to show g is analytic in D(0;1)

2. Relevant equations
Basic stuff about analytic functions and series representations, etc.

3. The attempt at a solution

Okay, so if z does not equal 0 then it is the quotient of two analytic functions (f is analytic by the hypothesis and any polynomial is analytic) hence g is analytic.

If z=0 then I wrote f as a Taylor series: f(z) = f(0) + f'(0)z + (f''(0)/2)*z^2 +...
so g'(0) = [f'(0)]' = [0]' = 0 thus g is differentiable at z=0; and I'm done.

Is my reasoning correct?