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I figured I can somehow use Liouville's theorem. I don't really know what the method is for solving or where to begin.

Suppose that $f: \mathbb{C} \to \mathbb{C}$ is an entire function. Let $g(z) = f(\frac{1}{z})$. Prove that if $g$ has a pole at $0$ then there is a $z_{0}$ in $\mathbb{C}$ such that $f(z_{0}) = 0$. Can assume $g$ is holomorphic on $\mathbb{C} \setminus \{0\}$.

José Carlos Santos
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Sam Bailey
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  • Please see https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Angina Seng Aug 18 '20 at 15:57
  • Can you prove that $f$ is a polynomial function? – Angina Seng Aug 18 '20 at 15:59
  • if f is entire then it can be represented as a power series. then i can use this result to show that is it a polynomial function of degree m https://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial

    then i can somehow use that result to show that its surjective or something

     – Sam Bailey Aug 18 '20 at 16:07

2 Answers2

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Since $f$ is entire, there is a power series $\sum_{n=0}^\infty a_nz^n$ centered at $0$ such that$$(\forall z\in\Bbb C):f(z)=\sum_{n=0}^\infty a_nz^n$$and therefore$$(\forall z\in\Bbb C\setminus\{0\}):g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$$So, since $g$ has a pole at $0$, there is some $N\in\Bbb N$ such that $n>N\implies a_n=0$. So, $f$ is a polynomial function and therefore you can use the Fundamental Theorem of Algebra to prove that it has a zero.

José Carlos Santos
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  • thanks i understand most of that. do you know where i can further learn about the line

    "So, since g has a pole at 0, there is some N∈N such that n>N⟹an=0" as i dont quite fully understand why

     – Sam Bailey Aug 18 '20 at 16:29
  • sorry i keep typing enter to write a new line but it posts the comment, i edited it to what i was going to type haha – Sam Bailey Aug 18 '20 at 16:30
  • @JoséCarlosSantos , i am not getting line ' So ,since $g$ has a pole at $0$..." can you please elaborate this? – TheStudent Aug 18 '20 at 16:31
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    You have$$g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$$and having a pole at $0$ means that the coefficients of $\frac1{z^n}$, with $n\in\Bbb N$, are $0$ when $n$ is large enough. That is what I have used here. – José Carlos Santos Aug 18 '20 at 16:32
  • Thank you for your response @JoséCarlosSantos, now i understand this. – TheStudent Aug 18 '20 at 16:35
  • thanks for the explanation! – Sam Bailey Aug 18 '20 at 16:35
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You will need the minimum modulus principle: If the modulus of a holomorphic function on a connected domain has a minimum, then either the function is constant or the minimum is $0$. In this case, you should show that:

a) $f$ is not constant.

b) There is a suitably large open disc $U_R(0)$ on which $\vert f\vert$ has a minimum.

 

For a), show that if $f$ were constant, then $g$ would have a removable singularity.

For b), take a compact disc $\overline{U_r(0)}$ and show that $\vert f\vert$ has a minimum $m$ on said disc. Then show that there exists another, larger compact disc $\overline{U_R(0)}$ where $\vert f(z)\vert>m$ for all $z\in\partial U_R(0)$. Conclude that $\vert f\vert$ has a minimum on the interior of $U_R(0)$. Then apply the minimum modulus principle.

Vercassivelaunos
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