Application of Argument Principle to Determine number of zeros

Question

Show that there is exactly $1$ real solution in the unit disk $D = \{ z : |z| < 1 \}$ to the equation $ze^{\lambda - z} = 1$, where $\lambda \in \mathbb{R} : \lambda > 1$. The fact that there is one real solution in the unit disk follows from the intermediate value theorem. That is $f(x) = xe^{\lambda -x} - 1$ is a continuous real valued function and $f(-1) = -e^{\lambda+1} -1 < 0$ and $f(1) = e^{\lambda - 1} - 1 > 0$. I am trying to consider the change in argument along the curve $\gamma = \{ e^{it} : t \in [0,2\pi] \}$ and show that the change in argument of $f(z)$ is $2\pi$, but I am not sure how argue that. I know $f(e^{it}) = e^{it} e^{\lambda - \cos(t)} e^{-i\sin(t)} - 1$. I want to say that the $e^{it} e^{\lambda - \cos(t)} e^{-i\sin(t)}$ dominates (as its modulus is greater than $1$), but I don't know how to prove it formally.

Answer 1

Try Rouché's theorem with $g(z) = ze^{\lambda -z}$. Since $|g(z)-f(z)| =1 < e^{\lambda-1} \le |g(z)|$ for $|z|=1$ we see that $g,f$ have the same number of zeros in $D$.