I'm supposed to use Rouche's theorem to solve this problem, but I'm pretty sure it's not possible. Can anyone confirm this? I want to determine how many zeros $e^z-z$ has on $B_1(0)$. The obvious set up is to take $f(z)=e^z$ and $g(z)=-z$, but Rouche's theorem can't be applied on $\partial B_1(0)$, as $e^{-1}<1$.
We are able to use Rouche on a smaller region. I believe $\partial B_{1/2}(0)$ works. The justification being that the modulus of $e^z$ should be minimized when $z=-1/2$, but $e^{-1/2}>1/2$. Then, we get that $e^z$ and $e^z-z$ have the same number of zeros in $B_{1/2}(0)$.
So $e^z-z$ has no zeros in $B_{1/2}(0)$. That's great, but not what the problem asked. Is there a tricky way to relate this to $B_1(0)$ somehow that I'm missing? This seems like it should be such a cut-and-dry application problem, but I'm just not seeing what to do. Is this even possible to do using Rouche?
Best Answer
For “small” $z$ is $e^z \approx 1 + z $ or $e^z - z \approx 1 \ne 0$. That suggests to apply Rouché's theorem to the functions $f(z) = e^z-z$ and $g(z) =1$: For $|z| = 1$ is $$ |f(z)-g(z)| = \left| \sum_{n=2}^\infty \frac{z^n}{n!}\right| \le \sum_{n=2}^\infty \frac{1}{n!} = e - 2 < 1 = |g(z)| $$ so that $f$ and $g$ have the same number of zeros in the unit disk, i.e. none.
Alternatively, use the triangle inequality instead of Rouché's theorem: For $|z| \le 1$ is $|e^z-1-z| \le e-2$ and therefore $$ |e^z-z| \ge 1 - |e^z-z-1| \ge 1 - (e-2) > 0 \, . $$