Question about a proof (Rouches Theorem) Context:
I made a video some time ago, where I tried to prove Rouches Theorem with the help of Argument Principle and Cauchy Integral Theorem. But now it seems like I made a mistake in the video which can have invalidated the proof and that is why I would like to research if the idea behind the proof is still valid or if you simply are not able to prove Rouches Theorem this way.
Prove: If f(z) and g(z) are analytic inside and on a simple closed curve C and if |g(z)| < |f(z)| on C, then f(z) + g(z) and f(z) have the same number of zeros inside C.
Proof:
We can describe the difference of the number of zeros (N) and the number of poles (P) for a function f(z) inside C with the help of the Argument Principle: $$N-P= \frac{1}{2\pi i}\oint_C \frac{f'}{f} \,dz $$
But for an analytic function the Argument Principle takes the following form since an analytic function have no poles ($P=0$): $$N= \frac{1}{2\pi i}\oint_C \frac{f'}{f} \,dz \quad (\text{if }f(z) \text{ is analytic})$$
This is the formula we will be using since the functions f(z), g(z) and f(z) + g(z) are analytic inside and on C from the information given above. The number of zeros for the two function (f(z) and f(z) + g(z)) can therefore be described as the following.
$$N_1 = \frac{1}{2\pi i}\oint_C \frac{f'(z)+g'(z)}{f(z)+g(z)} \,dz \qquad \text{and} \qquad N_2 = \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz$$
Note that $f(z) \not = 0$ and $f(z) + g(z) \not = 0$ since $|g(z)| < |f(z)|$ on C. If we are able to show that $N_1 - N_2 = 0$ it would imply that the two functions f(z) and f(z) + g(z) have the same number of zeros inside C which would complete the proof.
\begin{align}
N_1 - N_2 = \frac{1}{2\pi i}\oint_C \frac{f'(z)+g'(z)}{f(z)+g(z)} \,dz - \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz \quad (*)
\end{align}
To make the calculations a bit easier to read later on, let $F(z) = \frac{g(z)}{f(z)}$. This implies that $g'(z) = F'(z)f(z) + F(z)f'(z)$ from the chain rule and the fact that $g(z) = F(z)f(z)$, inserting this result into the calculations $(*)$ above yield:
\begin{align}
N_1 - N_2 &= \frac{1}{2\pi i}\oint_C \frac{f'(z)+F'(z)f(z)+F(z)f'(z)}{f(z)+F(z)f(z)} \,dz - \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz = \\
&= \frac{1}{2\pi i}\oint_C \frac{f'(z) (1+F(z))+F'(z)f(z)}{f(z) (1+F(z))} \,dz - \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz  = \\
&= \frac{1}{2\pi i}\oint_C \frac{f'(z) (1+F(z))}{f(z) (1+F(z))} \,dz + \frac{1}{2\pi i}\oint_C \frac{F'(z)f(z)}{f(z) (1+F(z))} \,dz - \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz = \\
&= \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz + \frac{1}{2\pi i}\oint_C \frac{F'(z)f(z)}{f(z) (1+F(z))} \,dz - \frac{1}{2\pi i}\oint_C \frac{f'(z)}{f(z)} \,dz = \\
&= \frac{1}{2\pi i}\oint_C \frac{F'(z)f(z)}{f(z) (1+F(z))} \,dz = \frac{1}{2\pi i}\oint_C \frac{F'(z)}{(1+F(z))} \,dz 
\end{align}
so if we can show that this last integral is equal to zero we are done, we might be able to do this with the Cauchy Integral Theorem, since if F'/(1+F) is analytic inside and on C then the integral must be equal to zero from the theorem. 
We know that $1 + F(z) = 1 + \frac{g(z)}{f(z)} \not = 0$ on C since $|g(z)| < |f(z)|$ on C, which means that F'/(1+F) is analytic on C. But we also need to show that F'/(1+F) is analytic inside of C since that would complete the proof. This last step (showing that it is analytic inside of C) is the thing I have not been able to prove which I describe a bit more below. Is it possible to show with the information given?
The Problem:
The source I used when I made the video was this one here since I thought that this proof was quite elegant and simple. But now I realized that this source did not motivate why F'/(1+F) is analytic inside the domain which is a crucial step in the proof since we need this information to be able to use Cauchy Integral Theorem which in turn completes the proof. The author just stated that this was the case (see step 3). I now continued by looking up the sources that the author referred to (Source_1 page 138 and Source_2 page 156) and the authors here are not even using the Cauchy integral theorem to prove the theorem.
Hence my question, can you use the Cauchy Integral Theorem and Argument Principle to prove Rouches Theorem like I tried to do in my video and here above or can we motivate why F'/(1+F) is analytic inside of the domain if it is analytic on the domain (which we have been able to prove) or from some other information since that would complete the proof.
Lastly:
Please let me know if you would like to have more information, if I missed anything crucial or if something is obvious wrong in the proof. I really would like to know if this way of proving it is simply wrong so I can remove the video if that is the case.
Edits
1 - Added the Prove and Proof section.
 A: The nicest proof I've seen, which proves an even stronger version, relies on the fact that $|1+z| < 1 +|z|$ describes the slitted plane where negative real axis is removed. This is because the inequality is equivalent to
$$
                   |1+z|^2 < 1 + 2|z|+|z|^2 \\
                      1+2\Re z +|z|^2 < 1 + 2|z|+|z|^2 \\
                      \Re z < |z|
$$
And $\Re z = |z|$ iff $\Re z \ge 0$ and $\Im z = 0$. So $|1+z| < 1 +|z|$ iff $z \in \mathbb{C}\setminus[0,\infty)$. This region is a classic domain of the logarithm. Because of this, any closed curve in this region has $0$ winding number around $0$.
Now suppose that $f,g$ are holomorphic on an simply-connected open region $\Omega$, and suppose $\gamma$ is a simple, closed, rectifiable curve in $\Omega$. Further suppose that $|f+g| < |f|+|g|$ on $\gamma$. Then $f/g$ is a closed curve in $\mathbb{C}\setminus[0,\infty)$ and, hence, its winding number around $0$ is $0$, which gives
$$
          0=\int_{(f/g)\circ\gamma}\frac{1}{z}dz \\
      = \int_{\gamma}\frac{1}{(f/g)}d(f/g) \\
      = \int_{\gamma}\frac{g}{f}\frac{gf'-g'f}{g^2}dz \\
      = \int_{\gamma}\frac{f'}{f}-\frac{g'}{g} dz
$$
Hence, the number of zeros of $f$ inside $\gamma$ must equal the number of zeros of $g$ inside $\gamma$, counted according to multiplicity.
