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The problem says: Using the Cauchy's integral formula show that if $f(z)$ is an entire function wich verifies $|f(z)|\leq1+2|z|^3$ $\forall z \in \mathbb{C}$ show taht $f(z)$ is a polynominal and calculate his grade.

It's obvious that is a polynominal because it is entire, so u can use Taylor theorem centered in 0 to show that $f(z)=\sum a_nz^n$ with $a_n=f^n(0)/n!$. But i don't know how to calculate the grade.

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Apply Cauchy's Integral Formula to write $f^{(n)}(0)$ as an integral over the circle of radius $R$ around $0$ and use an obvious bound for the integral to get $f^{(n)}(0)=0$ for $n>3$ by letting $R \to \infty $.

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  • $\begingroup$ I get that $f^n(0)=0$ for $3\leq n$ $\endgroup$ – Lecter Jun 26 '18 at 8:55
  • $\begingroup$ This cannot be true for $n=3$. $f(z)=z^{3}$ satisfies the hypothesis and $f^{(3)} \neq 0$ in this case. So check your estmates carefully. $\endgroup$ – Kavi Rama Murthy Jun 26 '18 at 8:58
  • $\begingroup$ I get that $|\displaystyle\int_{|z|=1}^{}\displaystyle\frac{f(z)}{z^{n+1}}dz|\leq{}\displaystyle\int_{|z|=1}^{}\displaystyle\frac{1}{|z|^{n+1}}dz+2\displaystyle\int_{|z|=1}^{}\displaystyle\frac{1}{|z|^{n-2}}dz$ and doing the both integrals, they get anulated $\forall{n\geq{3}}$. So where is my error? $\endgroup$ – Lecter Jun 26 '18 at 9:13
  • $\begingroup$ I have made a correction to my answer. $|\int_{\gamma } f(z)\, dz| \leq M L$ if $|f(z)| \leq M$, $L$ being the length of $\gamma $. Integrate over the circle of radius $R$, get an upper bound and let $R \to \infty $. In your estimate you are not getting $0$ for any derivative. $\endgroup$ – Kavi Rama Murthy Jun 26 '18 at 9:28
  • $\begingroup$ The thing is that (for z in the border of the R radius circle) $|\displaystyle\frac{f(z)}{z^{n+1}}|\leq{}\displaystyle\frac{1+2R^3}{R^{n+1}}$ and this tend to zero when R tend to infinity when $n\geq{3}$ $\endgroup$ – Lecter Jun 26 '18 at 9:44

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