If a non zero entire function has infinitely many zeros, does that always mean that it is transcendental? Converse is not true and we can see that $e^z$ is the counter example.
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2The zero function has infinitely many zeroes and is not transcendental... You probably meant to exclude this function. – Eric Towers May 19 '20 at 16:29
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Yes of course. Zero function is excluded – mathenthusiast May 19 '20 at 16:35
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Assume $P(f(z),z)\equiv 0$ for some non-zero polynomial $P(x,y)=\sum_{i,j}a_{i,j}x^iy^j$. Pick such $P$ of minimal degree. There are infinitely many $z_k$ where $f(z_k)=0$ and hence $P(0,z_k)=0$. It follows that $P(0,z)$ viewed as a polynomial of $z$ alone is the zero polynomial, i.e., $a_{0,j}=0$ for all $j$. Then $P(x,y)=xQ(x,y)$ with $Q$ of lesser degree. For all $z$ with $f(z)\ne 0$, we have $Q(f(z),z)=\frac1{f(z)}P(f(z),z)=0$, and by continuity (the roots of $f$ are isolated!) $Q(f(z),z)\equiv 0$, contradicting minimality of $\deg P$.
Hagen von Eitzen
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I am unable to understand the proof and its consequences. Does such a P always exist? And if so, what does it have to do with the transcendental nature of f? Please explain – mathenthusiast May 20 '20 at 04:40