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

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.

• A side remark: I have never seen the TeX \over command. Pretty cool 😎 – JustAnotherStackUser May 24 at 15:52
• $1 = |z| > \frac{e}{3} \geq \frac{e^{\Re z}}{3} =\frac{|e^z|}{3}$ when $|z| = 1$ – Jakobian May 24 at 15:52
• By Roche's theorem we need $|f|>|g|$ inside $\gamma$ so how does it help us? – J. Doe May 24 at 16:07
• @JDoe Not true, Rouché's theorem only requires $|f(z)|>|g(z)|$ on $\gamma$. – Adam Latosiński May 24 at 16:29

## 2 Answers

@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.

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

• If $W=W_0$ is the main branch solution, then $-W_0(-\frac13)\in(0,1)$. There is another real solution using the $-1$ branch $-W_{-1}(-\frac13)>1$. Now if you could show that all the other branches give also solutions outside the unit disk, then this would be an alternative solution. – LutzL May 26 at 7:43