If $0<|a|<1$ show that the equation $(z-1)^ne^z=a$ has exactly $n$ solutions in $\textrm{Re}(z)>0$.
Hint: consider $\gamma$ as the line segment from $iR$ to $-iR$ followed by the semicircle $|z|=R$, $-\pi/2\le\arg(z)\le\pi/2$, where $R>2$.
I try to use Rouche's Thm, by taking $f(z) = (z-1)^ne^z$ and $g(z) = a$, and then consider two cases, when $|z|=R$ and when $-\pi/2\le\arg(z)\le\pi/2$, both where $R>2$, and I want to prove that in both cases $|g(z)|<|f(z)|$, for the first case is fine, but for the second I don't know how to do it.
Maybe I am not using the hint in the correct way, could anyone help me please? Thanks!
I think that your approach is fine. Consider the entire functions $f(z)=(z-1)^n$ and $g(z)=ae^{-z}$ (a bit different from yours). Then $z\in \mathbb{C}$ is a solution of $(z-1)^ne^z=a$ iff $z$ is a zero of $f(z)-g(z)$.
In the half-plane $\text{Re}(z)\geq 0$ we have that $$|g(z)|=|a|e^{-\text{Re}(z)}\leq |a|<1 \tag{1}.$$ On the other hand for $\text{Re}(z)=0$, then $z=it$ with $t\in\mathbb{R}$ and $$|f(z)|=|(it-1)^n|=|\sqrt{1+t^2}|^n\geq 1 \tag{2}.$$ Moreover for $|z|= R\geq 2$, $$|f(z)|=|(z-1)^n|\geq (|z|-1)^n\geq (R-1)^n\geq 1 \tag{3}.$$ Now consider the closed curve $\gamma_R$ given in your hint. Then, for all $z$ along $\gamma_R$, $|g(z)|<1\leq |f(z)|\geq 1$ by (1), (2) and (3). Hence, by Rouche's Theorem, for any $R\geq 2$, $f(z)−g(z)$ has as many zeros as $f$ inside $\gamma_R$ (note that $f$ has $n$ zeros at $1$).