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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.

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

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

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  • $\begingroup$ 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. $\endgroup$ – LutzL May 26 at 7:43

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