# $f(z) = \frac{1}{z^2-2z+2}$ - $|f^{n}(0)| \leq n!$ - Cauchy inequality [closed]

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?

## closed as off-topic by heropup, Stefan Mesken, Claude Leibovici, user296602, JonMark PerryMar 24 '16 at 8:10

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• Do you know the formula $f^{(n)} (0)=\frac{n!}{2\pi i} \int_{C} \frac{f(z)} {z^{n+1}} dz$? – Mhenni Benghorbal Mar 21 '16 at 4:44
• You could compute $|f^n(0)|$ for the first few $n$ and then compare it to $n!$ then appeal to induction. Mhenni's idea is better though... – Nap D. Lover Mar 21 '16 at 4:47
• After I typed that answer I realized an even easier approach might be just doing partial fraction and get a closed form of $f^{(n)}(z)$, then plug in $0$. – user negative one over twelve Mar 21 '16 at 6:19

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 $c_n=\frac{1}{2}(b^{-n}+a^{-1}b^{-n+1}+...+a^{-n})=\frac{b^{-n-1}-a^{-n-1}}{2(b^{-1}-a^{-1})}=\frac{1}{2(\sqrt2)^n}\frac{\cos((n+1)\frac{\pi}{4})}{\cos(\frac{\pi}{4})}\leq \frac{1}{2\sqrt{2}^{n-1}}\leq 1$.