Bound for derivative of analytic function on a disk I have the following problem: Given a function analytic on a disk, with
$$|f(z)| \leq \frac{1}{(1-|z|)^n},$$
in the disk, for some positive integer $n$.
I want to show in that case, similar bound holds for derivative of $f$, i.e.
$$|f'(z)|\leq \frac{a(n)}{(1-|z|)^{n+1}},$$
where $a(n)\sim n$ (linear in $n$).
I am only able to show $a(n)\sim 2^n$ (exponential in $n$).
How should I approach this? Thank you for the help. 
 A: Use Cauchy's integral formula for the derivative:
$$f'(z)=\frac{1}{2\pi i}\int_{\Gamma_\rho}\frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta$$
where $\Gamma_\rho$ is a circle of radius $\rho$ centered at the origin, and $|z|<\rho<1$.
Parametrize $\Gamma_\rho$ using $\zeta=(z/|z|)\rho e^{i\theta}$, put $r=|z|$, and use the given estimate ($|f(\zeta)|\le 1/(1-\rho)^n$ on $\Gamma_\rho$) to obtain the estimate
$$|f'(z)|\le\frac\rho{2\pi(1-\rho)^n}\int_{-\pi}^\pi\frac{d\theta}{|\rho e^{i\theta}-r|^2}.$$
Now the classical substitution $u=\tan(\theta/2)$ yields $d\theta=2\,du/(1+u^2)$ and $\cos\theta=(1-u^2)/(1+u^2)$, and after subsituting these and simplifying we find
$$ \int\frac{d\theta}{|\rho e^{i\theta}-r|^2}=\int\frac{2\,du}{(\rho-r)^2+(\rho+r)^2u^2}=\frac2{\rho^2-r^2}\arctan v,\qquad u=\frac{\rho-r}{\rho+r}v.$$
Therefore we get
$$
|f'(z)|< \frac{\rho}{(1-\rho)^n(\rho^2-r^2)}=\frac\rho{\rho+r}\cdot\frac1{(1-\rho)^n(\rho-r)}.$$
Now put $\rho=r+(1-r)/n$, so $1-\rho=(1-r)(1-1/n)$, and get
$$|f'(z)|<\frac n{(1-r)^{n+1}(1-1/n)^n},$$
which ought to do the job.
