How to prove that the following function has exactly two zeroes in a particular domain? I am practicing exam questions for my own exam in complex analysis. This was one I couldn't figure out.
Let $U = \mathbb{C} \setminus \{ x \in \mathbb{R} : x \leq 0 \} $ en let $\log : U \to \mathbb{C} $ be the usual holomorphic branch of the logarithm on $U$ with $\log (1) = 0$. Consider the function given by $f(z) = \log(z) - 4(z-2)^2$.
Q1: Show that $f$ has exactly two zeroes (including multiplicity) in the open disk $D(2,1) = \{ z \in \mathbb{C} : |z-2| < 1 \} $.
Q2: Show that $f$ has exactly two different zeroes in $D(2,1)$.
I strongly suspect we should use Rouché's Theorem for this. I tried to apply it by setting $h(z) = \log(z)$ and $g(z) = -4 (z-2)^2$. If $|z-2| = 1$, then $z = 2 + e^{it}$ with $t \in [0,2 \pi ) $. Then we have $|h(z)| = |\log(z)| \leq \log|z| = \log|2+e^{it}| \leq \log(|2| + |e^{it}|) = \log|2|\cdot \log|e^{it}| = \log(2) \cdot \log(1) = 0$.
Furthermore, we have $g(z) = |-4 (z-2)^2| = |-4| |z-2|^2 = 4$. So $|h(z)| < |g(z)|$ when $|z-2| = 1$. According to Rouché's Theorem, $f$ en $g$ have the same number of zeroes within $\{ |z-2| < 1 \}$, so we have to count the zeros of $g$ to find the number of zeroes of $f$. However, I can only find one zero of $g$, which is at $z=2$. Can you tell what's going wrong with my approach?

Your estimate $|h(z)|\le 0$ (when $|z-2|=1$) cannot possibly be true: a nonconstant holomorphic  function cannot be equal to zero on a circle. I marked the incorrect steps in red:
$$|\log(z)| \color{red}{\leq} \log|z| = \log|2+e^{it}| \leq \log(|2| + |e^{it}|) \color{red}{=} \log|2|\cdot \log|e^{it}| = \log(2) \cdot \log(1) = 0$$
A correct estimate could look like
$$|\log(z)| \le |\operatorname{Re} \log z| + |\operatorname{Im} \log z| 
= \log |z| + |\arg z| \le \log 3+ \pi/2 <3$$ 
which is not sharp but suffices for the application of Rouché's theorem, which settles Q1. 
The question Q2 is less standard. One way to answer it is to observe that the real function $f(x)=\log x-4(x-2)^2$ has two distinct zeros on the interval $(1,3)$, because $f(1)<0$, $f(2)>0$, and $f(3)=\log 3-4<0$. Since we already know there are two zeros in $D(2,1)$ with multiplicity, there are no others.
