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Assume $f$ and $g$ are analytic on a domain $D$,

$\forall z \in \mathbb{C}$, $f(z)$, $g(z) \neq 0$.

Suppose that $|f(z)|=|g(z)|$ , $\forall z \in \partial D$.

Prove that $f(z) = Cg(z)$ $\forall z \in D$, while $|C|= 1$.

Would appreciate any kind of hints and tip :)

Second question, $$f(z) =\frac{1-\cos z}{z^2}$$ is entire function? I guess not because I believe $f(z)$ isn't analytic at $z = 0$, but how can I show that in a formal form?

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    Your "for all $z\in \mathbb C$" assumption makes no sense. $f,g$ are only defined in $D.$ Your second question should be removed, as it has nothing to do with the first question. – 2017-01-16

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Hint: use the maximum and minimum modulus principles for $f/g$. For your second question, yes it is entire: it has a removable singularity at $0$.

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    First of all, thanks for your comment! however i couldnt prove that with those principles because it seems that |C| wouldnt be equal to 1, i have tried assuimg f or g as a constant. sadly no success so far. – 2017-01-15
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    For example - |g(z)|=13 is constant on domain d. ∀z∈∂D |f(z)|=13, suppose |f(z0)|=5, z0∈D. in this case, |C|=5/13? where did i go wrong – 2017-01-15
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    @user407143 I think we may need to assume that $D$ is bounded. Are you sure that's not included in the hypotheses? – 2017-01-15
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    as a domain , D is bounded. yes it is included – 2017-01-15
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    @user407143 you shouldn't have a problem applying the two principles. You have to prove that $f/g$ is not constant on $D$, then it attains both its minimum and maximum on $\partial D$. Depending on what you took in class, you might just as well use this result directly. So you get a contradiction, and you conclude that $f/g=$ some constant $C$ on $D$. By continuity of $|f/g|$, $|f/g| = |C|$ on $\partial D$, in particular $|C| = 1$. – 2017-01-15