$Re(\frac{f}{g})\geq 0$ on the boundary of disk implies $f$ and $g$ have same number of zeros in the disk

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I came across the following problem while solving some question paper. I have not been able to make any progress on it.

“Let $f$ and $g$ be holomorphic in a neighbourhood of unit disk. Assume that $f$ and $g$ have no zeros on the boundary, and $Re(\frac{f}{g})\geq 0$ on the boundary. Show that $f$ and $g$ have same number of zeros in the disk.”

I do not understand how to proceed. I thought of using Rouche’s theorem but I do not know how to obtain $|f+g|<|f|+|g|$ on the boundary from the given condition.

I am not sure but I was also thinking if I could get from the above condition that $\frac{f}{g}$ extends as holomorphic function on the disk and then show that it must be a constant. But, I also feel that it would be a little strong thing to conclude as this would mean that $f$ and $g$ have the same zeros.

Any hint would be highly appreciated.

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The principal branch of $\log$ is analytic on $\Bbb{C}-(-\infty,0]$

$\frac{f(z)}{g(z)}$ is analytic and $ \in \Bbb{C}-(-\infty,0]$ for $|z| \in (1-\epsilon,1+\epsilon)$

thus $h(z)=\log \frac{f(z)}{g(z)}$ is analytic for $|z| \in (1-\epsilon,1+\epsilon)$ so that $$0=\int_{|z|=1} h'(z)dz=\int_{|z|=1} (\frac{f'(z)}{f(z)}-\frac{g'(z)}{g(z)})dz$$

The proof of Rouche's theorem is based on the same argument.