# A Continuity Argument in the Proof of Rouche's Theorem

In Greene and Krantz's Function Theory of One Complex Variable, the proof of Rouche's theorem involves the following continuity argument.

Let $f,g\colon U \to \mathbb{C}$ be holomorphic from an open set $U$. Let $\bar{D}(p,r) \subseteq U$, and $$\vert f(\zeta) - g(\zeta) \vert \le \vert f(\zeta) + g(\zeta) \vert$$ for $\zeta \in \partial D(p,r).$

Define $$f_t (\zeta) = tf(\zeta) + (1-t)\,g(\zeta),$$ for $t\in [0,1]$. Then the integral $$I_t = \frac{1}{2\pi i} \oint_{\partial D(p,r)} \frac{f'_t (\zeta)}{f_t (\zeta)}\,d\zeta$$ is a continuous function of $t\in [0,1]$.

They add in parenthases that the denominator does not vanish and the integrand depends continuously on $t$.

Question: Would anyone be kind enough to supply the details behind this continuity of $I_t$ ?

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Let $$F(t,\zeta)={tf'(\zeta)+(1-t)g'(\zeta)\over tf(\zeta)+(1-t)g(\zeta)}$$ Since the denominator does not vanish, the function $F$ is continuous on $[0,1]\times \partial D(p,r)$. This is a compact set, so $F$ is in fact uniformly continuous in $(t,\zeta)$.
Now, let $\epsilon>0$ be given, and let $t_0 \in [0,1]$. By the uniform continuity, there is some $\delta>0$ such that $$|F(t,\zeta)-F(t_0,\zeta)|<\epsilon/r$$for all $t$ with $|t-t_0| < \delta$. It follows that $$|I_t-I_{t_0}|\leq {1\over 2\pi}\Bigl|\int_{\partial D(p,r)}\bigl(F(t,\zeta)-F(t_0,\zeta)\bigr){\rm d}\zeta\Bigr| \leq {1\over 2\pi}\cdot {\epsilon\over r}\cdot 2\pi r=\epsilon$$ This means that $I_t$ is continuous with respect to $t$.