Prove that for every $w \in \mathbb{C}$ it is true that $z + \sin z = w$ has exactly one root $z \in \mathbb{C}$. Prove that for every $w \in \mathbb{C}$ it is true that $z + \sin z = w$ has exactly one root $z \in \mathbb{C}$. 
The idea I have is to use Rouche's theorem. 

This is not true. It says that the function $f(z)=z+\sin(z)$ is a bijection from $\mathbb C$ to itself; in fact it's fairly well known and not hard to prove that the only holomorphic bijections of the plane are of the form $az+b$.
(Say $g\in H(\mathbb C)$. If $g$ is not a polynomial then it has an essential singularity at $\infty$, so Weierstrass tells us it cannot be injective, since the image of every neighborhood of $\infty$ is open and dense. Otoh if $g$ is a polynomial it cannot be injective if the degree is greater than $1$. So in fact the only injective entire functions are of the form $az+b$.)
Or in yet more detail: Say $g$ has an essential singularity, say at $0$. Say $0<r<|p|$, and let $V=g(D(p,r))$. Then $V$ is a nonempty open set. Let $W=g(\{z:0<|z|<|p|-r\})$. Then Weierstrass says $W$ is dense. Hence $W\cap V\ne\emptyset$, so $g$ is not injective.
