The functions $f_{k}(x) = \sin(kx)$ are infinitely differentiable on $[-1,1]$ and bounded by 1. However their derivatives at 0 are unbounded.
If we have functions $f_{k}(z) = \sin(kz)$ on the unit disc, the Cauchy Estimates provide bounds for $(\frac{\partial}{\partial z }) f_{k}(0)$. Why are these examples not contradictory?
Well I am not sure, but what I think is that the z's are complex numbers and they have an imaginary component. However, on the real numbers, we are looking at functions from two dimensions; so there is only one direction. I am not sure. Try not to work out this problem completely. Thanks guys!!!
The Cauchy estimates provide bounds for $f_k'(0)$ for each $k\in\mathbb N$. There is no contradiction between that and the assertion that the set $\{f_k'(0)\mid k\in\mathbb N\}$ is unbounded.
Besides, you wrote that $f_k$ is bounded by $1$. If that means that $(\forall z\in\mathbb C):\bigl\lvert\sin(kz)\bigr\rvert\leqslant1$, then you are wrong. In fact, no $f_k$ is a bounded function.