Let a be a complex number with $Re(a)>1$. How many solutions exist for the equation $e^z-z=a$ at $Re(z)<0$.
I assume this is an exercise on Rouché's theorem, but this is the first time I see this kind of exercise, so I don't really understand how to implement the theorem to solve this. Any help would be greatly appreciated.
Let $1 < r < \operatorname{Re}(a)$ and $\gamma = \{z \mid \lvert z + a \rvert = r \}$. Then $\lvert \exp(z) \rvert < 1 < \lvert z+a \rvert$ for $z \in \gamma$. Note that any solution of your equation in the left half plane must lie in the interior of $\gamma$. Now apply Rouché to $z + a - \exp(z)$ and $z+a$.
Alternative approach: Let $D$ denote the closed unit disc centered at $0$. Note that $w \mapsto \exp(w-a)$ maps $D$ into itself and is a contraction on $D$ (its derivative has norm strictly smaller than $1$ on $D$). Therefore there exists exactly one $z_0$ in $D$ such that $\exp(z_0-a) = z_0$.
Substitute $z_0 \leftarrow z + a$ to conclude that $\exp(z)-z=a$ has exactly one solution in the closed disc $D-a$ (centered at $-a$). However, for any $z$ in the left half plane not in this disc $\lvert \exp(z) \rvert < 1 < \lvert z+a \rvert$. This sows that there is exactly one solution of your equation in the left half plane.