Complex Analysis: Showing abs{f(z)} ≤ abs{z^k}

[esisk]

1. If f(z) : D--->D is analytic where D is the open unit disk, and

the first (k-1) derivatives at zero vanish i.e (f(0)=0,f'(0)=0,f''(0)=0...f^k-1(0)=0



2.I would like to show that

abs{f(z)} \leq abs{z^k}





3. I believe one can (an the question is possibly intended to be solved this way) approach it using Schwarz, I was trying to do it by brute force whereby expanding the power series and saying
f(x)=z^k g(z) and somehow show that I g(z) I <= 1. I am failing as of now.
Any help? I thank you for your time

 

[Dick]

g(z)=f(z)/z^k is analytic in the disk, right? Look at the limit as |z|->1 and apply the maximum modulus theorem, just like in the proof of the Schwarz lemma. I don't see how you can 'brute force' the power series. You need to use that f:D->D.

 

[esisk]

I thank you for the response Dick...I appreciate the hint, I am working on it. I know it is elementary, but I am having difficulty.