How to discard incorrect options??

Question

Suppose $f$ and $g$ are entire functions $g(z)\neq 0 \forall z\in \mathbb C$.If $\vert f(z)\vert \le \vert g(z)\vert $,then we can conclude that-

• $f(z)\neq 0 \forall z\in C$
• $f$ is a constant function.
• $f(0)=0$
• for some $M\in \mathbb C$,$f(z)=Mg(z)$.

Solution:

Since,$f(z)$ and $g(z)$ are entire functions and $g(z)\neq 0\implies \frac{f(z)}{g(z)}$ is also entire.

Now,$\vert f(z)\vert \le \vert g(z)\vert \implies \frac{\vert f(z)\vert}{\vert g(z)\vert} \le 1. \tag{*}$So,by Liouville's theorem $f/g$ is constant.

Making $4$ option true.

Now,since $g(z)\neq 0 z\forall C$ and is entire so,$g(z)$ must be a non-zero constant.

Thus, by $f/g$ is constant ang $g $ is non-zero constant we'll get $f$ is constant..Making option $2$ correct.

But the only correct option is $4$.

How to discard options 1,2,3??

Compilation of arguments suggested by José Carlos Santos and kamills

• Since,$f(z)=Mg(z)$.But,$M$ can be equal to zero.So,$f(z)=0$,making option $1$ false.
• Taking,$f(z)=e^z;g(z)=1+e^z$,then $\vert f(z) \vert \le \vert g(z) \vert$.But,$f(z)$ is not constant,making option $2$ false.
• take $f(z)=e^z$,then $f(0)=1$,making option $3$ false.

Answer 1

You can't discard the second or the third options. After all, your $M$ can very well be equal to $0$. And, by the way, that proves that you can discard the first option.

Answer 2

I'm not completely convinced that $g$ has to be constant. After all, the hypothesis is only that $|f(z)| \leq |g(z)|$ on $\mathbb{C}$; it's not assumed that $g$ is bounded. Although it is entire, Liouville's Theorem can't be used, so option 2 isn't necessarily correct.

Option 1 can also be discarded, since even though $g$ isn't zero, $f$ could be at some point of $\mathbb{C}$. Similarly, there's no reason to believe that $f(0) = 0$, discarding option 3.