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Show that a map $f\in C^1(\mathbb{C})$ with $|f(z)|\leq C |z|^{\alpha}$ for some $a<n+1$, a constant $C>0$ and all $z\in\mathbb{C}$ is a polynomial of a degree less or equal to $n$.


This was given to me as a hard bonus exercise as part of my real analysis/multivariable calc course (first year student). Sadly, too hard for me and we only touched upon very introductionary complex analysis via Taylor series etc. I'm not even seeing the right approach. I'd be very thankful for somebody illustrating to me how to proceed here. I've never seen a proof for an exercise of this kind before.

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    $\begingroup$ How about $f(z)=\frac{1}{1+|z|^2}$? Apparently, you meant, $f$ entire analytic, satisfying the inequality. $\endgroup$ – Yiorgos S. Smyrlis Jul 2 '18 at 11:37
  • $\begingroup$ It is quite straightforward from the complex-analytic viewpoint (assuming that $f$ is holomorphic). See this math.stackexchange.com/questions/438310/… I believe you can justify the gaps and mentioned results using just multivariable-calculus $\endgroup$ – Alan Muniz Jul 2 '18 at 11:41
  • $\begingroup$ @YiorgosS.Smyrlis $|z|^2 = Re(z)^2+Im(z)^2 = x^2+y^2$. CR equations should imply $U_x(x,y) = 2x = V_y(x,y) = 0$. So $x=0$. Using the other equation you get that CR equations are only satisfied in $z=0$. So $|z|^2$ isn't analytic in any open set of $\mathbb{C}$. $\endgroup$ – Rafael Gonzalez Lopez Jul 2 '18 at 14:44
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    $\begingroup$ $C^1(\mathbb C)$ is odd notation; what does it mean exactly? $\endgroup$ – zhw. Jul 2 '18 at 15:10

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