$f(z) = \frac{1}{z^2-2z+2}$ - $|f^{n}(0)| \leq n!$ - Cauchy inequality Let the function $f(z) = \frac{1}{z^2-2z+2}$. Show that for each integer $n \geq 0$, we have $|f^{n}(0)| \leq n!$.
Honnestly, I don't know how to do that problem, maybe in using the Cauchy inequality (complex analysis).
Is anyone could help me at this point?

We'll use the following result from geometric series to compute the power series of $f$.

Let $c\neq 0$, the power series of $\frac{1}{z-c}$ around $0$ is $\frac{1}{z-c}=\frac{\frac{-1}{c}}{1-\frac{z}{c}}=\frac{-1}{c} (1+(\frac{z}{c})+(\frac{z}{c})^2+(\frac{z}{c})^3+...)$, radius of convergence is $c$.

$f(z)=\frac{1}{(z-a)(z-b)}$, $a=1+i=\sqrt{2}e^{\pi i/4}$, $b=1-i=\sqrt{2}e^{-\pi i/4}$
$$f(z)\\=\frac{1}{z-a}\frac{1}{z-b}\\=\frac{-1}{a}\frac{-1}{b}(1+(\frac{z}{a})+(\frac{z}{a})^2+(\frac{z}{a})^3+...)(1+(\frac{z}{b})+(\frac{z}{b})^2+(\frac{z}{b})^3+...)\\=\frac{1}{2}(1+(\frac{z}{a})+(\frac{z}{a})^2+(\frac{z}{a})^3+...)(1+(\frac{z}{b})+(\frac{z}{b})^2+(\frac{z}{b})^3+...)$$
Because $\frac{f^{(n)}(0)}{n!}=c_n=$ the coefficient in front of $z^n$ in the expansion of $f(z)$. It's sufficient to prove $|c_n|\leq 1$ for each $n$.
Not hard to see from the above equation $c_n=\frac{1}{2}(b^{-n}+a^{-1}b^{-n+1}+...+a^{-n})=\frac{b^{-n-1}-a^{-n-1}}{2(b^{-1}-a^{-1})}=\frac{1}{2(\sqrt2)^n}\frac{\cos((n+1)\frac{\pi}{4})}{\cos(\frac{\pi}{4})}\leq \frac{1}{2\sqrt{2}^{n-1}}\leq 1$.
