I am a Mathematics student and I am studying complex analysis.I encountered a problem which is as follows:
Problem:
State whether true or false: An entire function $f:\mathbb{C\to C}$ satisfying $f(z)\to 0$ as $z\to \infty$ is a constant.
I think the statement is true.My reasoning goes in the following way:
for $\epsilon>0$ there exists $R>0$ such that $|f(z)|\leq\epsilon$ for all $|z|>R$.
Now consider the compact disc $\overline{D(0,R)}$,since $f$ is entire,hence $f$ is continuous so bounded on this compact set and bounded by $\epsilon$ outside the circle,so it is a bounded function,hence it is a constant.
Is my reasoning correct?
Your argumentation is correct, however I would still like to show another approach to solve this problem:
Since $f$ is entire, it is given by an everywhere converging power series in $0$, so: $f(z)= \sum_{n=0}^\infty a_n z^n$ on $\mathbb{C}$. Now we can see that the laurent series of $f(\frac{1}{z})$ is given by: $\sum_{n=0}^\infty a_n (\frac{1}{z})^n$ on $\mathbb{C}\setminus\{0\}$, since the laurent series is unique.
Now since $f$ tends to $0$ as $z \rightarrow \infty$, we have that $f(\frac{1}{z})$ tends to $0$ as $z \rightarrow 0$. Therefore, by the riemann removable singularity theorem we have that $f(\frac{1}{z})$ has got a removable singularity at $0$, therefore its principle part has to vanish. So we can conclude that: $f(\frac{1}{z})= \sum_{n=0}^0 a_n (\frac{1}{z})^n = a_0$. But since $f(\frac{1}{z})$ is constant, so must be $f$.