# Complex analysis

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