Find the number of zeros of $f(z)={1\over3}e^z-z$ in the unit disc

Question

Find the number of zeros of $f(z)={1\over3}e^z-z$ in the unit disc.

The book's solution is that because $|z|>{1\over3}e^z$ then by Roche's theorem $\mathbb{Z}(f)=1$, but I think it's a mistake since, for example, for $z=1$, $1>{e\over3}$ but for $z=0, 0\ngtr{1\over3}$. I'm not sure how to solve it, thanks.

Answer 1

@Jacobian in his comment above has already solved the problem. First read Rouche’s theorem carefully: Let $D$ be a bounded domain with piecewise smooth boundary $\partial D$. Let $f(z),h(z)$ be analytic in $D\cup \partial D$. If $|h(z)|<|f(z)|$ for all $z\in \partial D$, then $f$ and $f+g$ have the same number of zeros inside $D$.

Note than the image of your curve $\gamma$ equals $\partial D$ for some domain $D$, so inside $\gamma$ means for $z\in \partial D$ as in the theorem.

Answer 2

May be one of the precise root is -W(-$1\over3$), where $W$ is Lambert W function