Let $f:\mathbb{C}\to\mathbb{C}$ and $g:\mathbb{C}\to\mathbb{C}$ be analytical functions on all $|z|\leq R$, such that $|f(z)|=|g(z)|$ for all $|z|=R$ and $f,g\neq 0$ for all $|z|<R$, prove that there exists $\lambda\in\mathbb{C}$ such that $|\lambda|=1,\ f(z)=\lambda g(z)$, on the disk.
I defined de auxilliary function $h(z):=\frac{f(z)}{g(z)}$ and assumed that $|h(z)|=1$ for all $|z|=R$ and that the function is also analytic (or at least continuous) on $|z|\leq R$, then using the maximum modulum principle on both $h$ and $\frac 1h$, I got that $h$ is constant with modulus 1, thus proving the desired theorem.
The thing is that if $h$ is not well behaved on the circle I can't make that assumption. I know that there exists at least a neighbourhood of a point $z_0$ on the curve where $|h(z_0)|=1$ and it is analyitic. This because there exists at least a point (and therefore a neighbourhood from this point) on the circle where $|f(z_0)|=|g(z_0)|>0$ as there must be a point on the curve where its modulus is greater than all the points inside the curve. How do I conclude from this that I can make de assumptions made on the first paragraph? And if I can't then how can I prove the desired theorem?
I am assuming that $f,g$ are analytic on an open set containing the disk $|z| \le R$.
Note that $f,g$ are non zero on $D=B(0,R)$ hence $h={f \over g}$ and $j = {g \over f}$ are analytic on $D$.
Since $f,g$ are non zero, they are non zero on $\partial D$ (maximum modulus theorem). Furthermore, by the uniqueness theorem, any zeros of $f,g$ on $\partial D$ must be isolated. Furthermore, the orders of the zeros of $f,g$ must be the same otherwise we could not have $|f(z)|=|g(z)|$ nearby. Consequently, for any zero $z_0 \in \partial D$, we have that the limit $\lim_{z \to z_0} {f(z) \over g(z)}$ exists and is non zero. Hence $h$ has a removable discontinuity at $z_0$. The same analysis applies, mutatis mutandis, to $j$.
In particular, $|h(z)| = |j(z)| = 1$ for $z \in \partial D$.
By the maximum modulus theorem we see that $|h| \le 1$ and $|j| \le 1$ on $D$ hence $|h| = 1$ on $D$.
By the open mapping theorem, $h(U)$ is open if $h$ is non constant.
Since $S=\{z | |z| = 1 \}$ contains no open set and $h(U) \subset S$ we must have $h$ is a constant on $U$. Since $U$ is connected, there is some $\lambda$ with $|\lambda| = 1$ such that $h(z) = \lambda$ for all $z \in U$.