This particular question was asked in quiz yesterday and I was unable to solve it.
Suppose f and g are entire functions and g(z) $\neq$0 for all z $\epsilon \mathbb{C} $ . If |f(z) |$\leq$ |g(z) | , the which one of them is true.
1.f is a constant function.
2.f(0) =0 .
3.for some C $\epsilon \mathbb{C} $ , f(z) = C g(z) .
4.f(z) $\neq$ 0 for all z $\epsilon \mathbb{C} $ .
4 th option can be removed by taking f(z) =0 and g(z) =2 .
But I am unable to find any way to remove rest of options.
Consider $h(z)=\frac{f(z)}{g(z)}$ which is well defined because $g(z)\neq0$, then $h$ is an entire function and for all $z\in\mathbb{C}$ $|h(z)|\leqslant 1$, then by Liouville's theorem $h$ is a constant function, which holds 3.
Simply consider the function $z \mapsto \frac{f(z)}{g(z)}$. Since, $g(z) \ne 0, \forall z \in \Bbb C$ we have that this map is entire and as $|f(z)| \le |g(z)|$ we have that our map $z \mapsto \frac{f(z)}{g(z)}$ has absolute value bounded by 1 $\forall z \in \Bbb C$, hence Liouville's theorem gives (3).
(1) and (2) aren't true by taking $f(z)=g(z)=e^z$
(4) isn't true by taking $f \equiv 0$ and $g \equiv 1$
