Tricky Application of Rouche's Theorem

Question

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?

Answer 1

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.

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Alternatively, use the triangle inequality instead of Rouché's theorem: For $|z| \le 1$ is $|e^z-1-z| \le e-2$ and therefore $$ |e^z-z| \ge 1 - |e^z-z-1| \ge 1 - (e-2) > 0 \, . $$

Answer 2

You can try the symmetric version of Rouché':

If $f$ and $g$ are analytic in a neighbourhood of $K$ and $|f(z) - g(z)| < |f(z)| + |g(z)|$ for $z \in \partial K$, then $f$ and $g$ have the same number of zeros in $K$.

Note that $|f(z) - g(z)| \le |f(z)| + |g(z)|$ with equality only if $f(z)$ and $g(z)$ are on opposite sides of the same line through the origin.

So take $f(z) = \exp(z) - z$ and $g(z) = \exp(z)$. On the unit circle $\partial K$ we have $|f(z) - g(z)| = 1$. If $z = x + i y \in \partial K$ with $x > 0$, $|g(z)| = \exp(x) > 1$ so $|f(z)| +|g(z)| > 1$. On the other hand, if $x \le 0$,
$\text{Re}(g(z)) = \exp(x) \cos(y) > 0$ and $\text{Re}(f(z)) > -x \ge 0$, so $f(z)$ and $g(z)$ can't be on opposite sides of the same line through the origin.