Limit of the quotient of two analytic functions $|f(z)|=|g(z)|$

Question

Let $f,g$ be two analytic functions on the domain $\Omega$ such that $|f(z)|=|g(z)|$ throughout $\Omega$.

I believe $h(z)=f/g$ only has removable singularities(can't really prove it...), for the following reasons. If $g(z_0)=0$, then $f(z_0)=0$, and $$\lim_{z\to z_0}h(z)=\lim_{z\to z_0}\frac{f(z)}{g(z)}=\lim_{z\to z_0}\frac{|f(z)|e^{\arg f(z)}}{|f(z)|e^{\arg g(z)}}\\ =\lim_{z\to z_0}\frac{e^{\arg f(z)}}{e^{\arg g(z)}}=e^{\arg f(z_0)-\arg g(z_0)}.$$ So, we define $h(z_0)$ to be this value(EDIT this value is undefined :( ). Also, $$ \lim_{z\to z_0}h'(z)=\lim_{z\to z_0}\frac{(f'g-g'f)(z)}{g(z)^2}\\ =\lim_{z\to z_0}(\frac{f'}{g}-\frac{g'}{g}\cdot\frac{f}{g})\\ =\lim_{z\to z_0}\frac{f'-hg'}{g}=\ldots? $$ Now I cannot proceed to prove that $h'(z)$ exist at $z=z_0$.

How can I make $h$ analytic?

PS: if $h$ is made analytic, I can prove by integration that $f(z)=e^{\alpha i}g(z)$ for some fixed $\alpha\in \mathbb R$.

Any help with the problem?

Answer 1

The problem with your approach is that $\arg f(z_0) = \arg 0$ and $\arg g(z_0) = \arg 0$ are not defined, and actually $w \mapsto \arg w$ cannot be defined as a continuous function in the neighborhood of $w=0$.

But your assumption that $h = f/g$ has only removable singularities correct. It follows directly from Riemann's theorem on removable singularities because $h$ is bounded. Therefore $h$ can be extended to a holomorphic function on $\Omega$.

Finally, $|h(z)| \equiv 1$ implies that $h$ is constant because of the maximum modulus (or open mapping) theorem.

Answer 2

If $f$ has a zero of order $n$ at $z_0$ then $f(z)=(z-z_0)^{n}h(z)$ with $h$ analytic and non-zero in a neighborhood of $z_0$. This implies that $g$ also has a zero of order $n$ at $z_0$. Similarly, if $g$ has a zero of order $n$ at $z_0$ so does $f$. Hence the zeros of $g$ are cancelled by those of $f$ in the ratio $\frac f g$. This makes $\frac f g$ analytic function with modulus $1$ hence a constant (by MMP). The constant must have modulus $1$ so it is of the form $e^{i\alpha}$ where $\alpha$ is real.