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Does any one how to prove that every entire algebraic function is a polynomial? I'm under the impression that this can be achieved by showing that an algebraic function grows no faster than a polynomial.

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    Yes, suppose $f(z)$ is an entire function with polynomial growth, i.e. there are constants $C$ and $N$ such that |f(z)| < C |z|^N for all sufficiently large $|z|$. Then $f(z)/z^N$ has a removable singularity at $\infty$, so $f(z)$ has at most a pole there, and so $f(z)$ is meromorphic on the Riemann sphere. But then $f(z)$ is a rational function, and since it has no poles in $\mathbb C$ it is a polynomial.2012-02-02

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I assume you mean every entire algebraic function is a polynomial. Suppose $g$ is an entire algebraic function of order $n$. Thus there are polynomials $c_j(z)$, $j=0 \ldots, n$ with $c_n$ not identically $0$ such that $\sum_{j=0}^n c_j(z) g(z)^j = 0$. For all but finitely many complex numbers $w$, $\sum_{j=0}^n c_j(z) w^j$ is not identically $0$, and so there are at most finitely many $z$ for which $\sum_{j=0}^n c_j(z) w^j = 0$: those are the only $z$ for which we can have $g(z) = w$. Thus for all but finitely many $w$, $g(z)$ takes the value $w$ only finitely many times. But an entire function that is not a polynomial has an essential singularity at $\infty$, and by the Great Picard Theorem it takes all but one value infinitely many times.

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    @Ger, it worked. Thanks.2012-02-02
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Hopefully not, because $\: \operatorname{exp} : \mathbb{C} \to \mathbb{C} \:$ defined by $\:\:\:\: f(z) \:\: = \:\: \displaystyle\sum_{n=0}^{\infty} \: \left(\frac1{n!} \cdot \left(z^n\right)\right)$
is an entire function that is not a polynomial.


See http://en.wikipedia.org/wiki/Exponential_function#Complex_plane.

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    @RickyMaher: "$\frac{1}{|z|}|f|$" is not a statement. What do you mean by "$\frac{1}{|z|}|f|$ holds"?2012-02-02