Roots of $e^z – z – 1$ on inside the unit circle

Question

I want to use Rouche's theorem to determine the number of zeros of $e^z – z – 1$ in the unit disk, but I cannot seem to get the right functions to apply Rouche's.

Answer 1

You can apply Rouché's theorem to $f(z) = e^z-1-z$ and $g(z) =z^2/2$: For $|z| = 1$ is $$ |f(z)-g(z)| = \left| \sum_{n=3}^\infty \frac{z^n}{n!}\right| \le \sum_{n=3}^\infty \frac{1}{n!} = e - \frac 52 < \frac 12 = |g(z)| $$ so that $f$ and $g$ have the same number of zeros in the unit disk.

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But Rouché's theorem is actually not needed here, only the triangle inequality: $$ f(z) = \sum_{n=2}^\infty \frac{z^n}{n!} = z^2 \left( \frac12 + \sum_{n=3}^\infty \frac{z^{n-2}}{n!}\right) $$ has a double zero at the origin, and the second factor has no zero in the unit disk since $$ \left|\sum_{n=3}^\infty \frac{z^{n-2}}{n!}\right| \le \sum_{n=3}^\infty \frac{1}{n!} = e - \frac 52 < \frac 12 $$ for $|z| \le 1$.