Let, $f,g$ be two holomorphic functions in a region $\Omega$. $C$ be a circle in $\Omega$ containing interior such that $|f(z)|>|g(z)|\ \forall z\in C$. g vanishes nowhere on $C$. Then $f$ and $f+g$ have same number of zeros inside $C$.
In the proof, we construct for each $t\in[0,1]$, $f_t=f+tg$. Then $f_t$ is non vanishing on $C$ and holomorphic on $\Omega$. So, by Argument principle ${1\over2\pi i}\int_{C}{f_t'(z)\over f_t(z)}dz=n_t=$#of zeros of $f_t$ inside $C$.
Now, we want to prove that ${f_t'(z)\over f_t(z)}$ is jointly continuous (hence uniformly continuous) on $[0,1]\times C$. If we can prove that $f_t$ is jointly continuous on $[0,1]\times C$, then $f'_t$ will follow similarly.
But I faced to prove the continuity.
Let $(t_0,z_0)\in[0,1]\times C$, choose $\varepsilon>0$, $|f_t(z)-f_{t_0}(z_0)|=|f(z)-f(z_0)+tg(z)-t_0 g(z_0)|$. We need to find a $\delta>0$ such the expression is less than $\varepsilon$ for $\sqrt{(t-t_0)^2+|z-z_0|^2}<\delta$
How to find such $\delta$ and establish the continuity?
Thanks for assistance in advance.