Assume that $g=x+iy$ be an entire function.
By a theorem(*), $\vert x(z)\vert \le N \vert z\vert ^n \ \ \forall z$ large enough and for constant $N\gt 0$ and for non-negative $n\in \Bbb Z$ $\Rightarrow$ $g$ is a polynomial of degree at most $n$.
Accourding to the proof (**), $g$ is C-analytic in $D(0,M$), and $M\gt 0$ $\Rightarrow$ $$g(z)=\frac{1}{2\pi} \int_{0}^{2\pi}x(Me^{i\theta})\frac{Me^{i\theta}+z}{Me^{i\theta}-z}d\theta +iy(0)$$
Supposing differentiation w.r.t $z$ can be carried inside the integral sign $\forall$ orders, give an alternative proof of theorem(*) by proving that $g^{(n+1)}(z_0)=0$ $\forall z_0\in \Bbb C$
I dont really have an idea. Please help me solving this. Thank you lot.
Note that $$\frac{d}{dz}\left(x(Me^{i\theta})\frac{Me^{i\theta}+z}{Me^{i\theta}-z}\right) = \frac{2Me^{i\theta}x(Me^{i\theta})}{(Me^{i\theta}-z)^2},$$ so we can deduce that for all $n>1$, $$\left(x(Me^{i\theta})\frac{Me^{i\theta}+z}{Me^{i\theta}-z}\right)^{(n)} = n!\frac{2Me^{i\theta}x(Me^{i\theta})}{(Me^{i\theta}-z)^{n+1}}.$$ Use the fact that $|x(z)|<|g(z)|<N|z|^n$ for large enough $z$ (and the fact that $g$ is complete) to bound $$|g^{(n+1)}(z)|\leq |y(0)| + \frac{1}{2\pi}\int_0^{2\pi}(n+1)!2M\frac{|x(Me^{i\theta})|}{|Me^{i\theta}-z|^{n+2}}d\theta \leq\\ \leq |y(0)| + \frac{1}{2\pi}\int_0^{2\pi}(n+1)!2M\frac{NM^n}{(M - |z|)^{n+2}}d\theta = |y(0)| + (n+1)!2M\frac{NM^n}{(M - |z|)^{n+2}},$$ and the second term tends to zero as $M\to\infty$, proving at least (by Liouville) that $g^{(n+1)}$ is constant, hence $g$ is a polynomial of degree at most $n+1$. I'm having difficulty right now tightening the bound to show it's actually of degree $n$, so perhaps someone else could carry us home?