Tricky Application of Rouche's Theorem

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?

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 $$|e^z-z| \ge 1 - |e^z-z-1| \ge 1 - (e-2) > 0 \, .$$

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.