Q:

Show there is an unique complex number $z \in D(0,1)$ such that $ze^{\lambda-z}=1$

I need to show there is an unique complex number $z \in D(0,1)$ such that $ze^{\lambda-z}=1$ where $\lambda>1$.
I know already, from the intermediate value theorem, there is an $z \in ]0,1[$ that satisfies this.
How do I prove this z is unique in $D(0,1) \subset \mathbb{C} $?
Thank you!

A:

Consider $f(z)= -ze^{\lambda}$ and $g(z)=e^z$. Since $\lambda>1$
$$|g(z)|=|e^z|<e^{|z|}=e<e^{\lambda}=|f(z)| \quad \forall z \quad \text{with} \quad |z|=1$$
So by Rouche's theorem $f$ and $g+f$ has the same number of roots. $f$  has only one root $z=0$, so $f+g$ it does too.

