Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am supposed to prove that $$\frac{|e^z-1|}{e-1} \leq |z|$$ for $|z| \leq 1$. My guess is that I have to show that the LHS $\leq 1$ and then apply Schwarz's Lemma. But I am not able to prove that!

share|cite|improve this question
up vote 11 down vote accepted

We can write \begin{align} |e^z-1|&=\left|\sum_{n\geq 1}\frac{z^n}{n!}\right|\\ &\leq \sum_{n\geq 1}\frac{|z|^n}{n!}\\ &\leq |z|\sum_{n\geq 1}\frac 1{n!}\\ &=|z|(e-1) \end{align}

share|cite|improve this answer

Try $\mathrm e^z-1=z\int\limits_0^1\mathrm e^{tz}\,\mathrm dt$. Hence $|\mathrm e^z-1|\leqslant|z|\int\limits_0^1|\mathrm e^{tz}|\,\mathrm dt$. Since $|\mathrm e^{tz}|=\mathrm e^{t\Re z}\leqslant\mathrm e^{t|z|}\leqslant\mathrm e^t$, the integral is at most $\int\limits_0^1\mathrm e^{t}\,\mathrm dt=\mathrm e-1$, hence $|\mathrm e^z-1|\leqslant|z|\cdot(\mathrm e-1)$, as desired.

This method proves more generally that, for every $z$, $|\mathrm e^z-1|\leqslant|z|\cdot(\mathrm e^{|z|}-1)$.

share|cite|improve this answer

Define $f(z) = \frac{e^z - 1}{e - 1}$. This is a holomorphic function, and it satisfies $f(0) = 0$. Thus, by Schwarz's Lemma, $|f(z)| \leq |z|$ for all $|z| < 1$.

Now all that remains is extending this result to $|z| \leq 1$. This is a simple technical result: The function $g(z) = |f(z)| - |z|$ is a continuous, real, non-positive* function defined in the entire complex plane. By continuity, it cannot attain a positive value at a point on the unit circle; otherwise, it would have to be positive at some neighborhood of that point, which includes points inside the disk.

*edit: I meant non-positive when $|z|<1$ as we've shown.

share|cite|improve this answer
Hi Yoni! But you have to show that the function $f(z)$ maps into $D$ (the open unit disk) in order to apply Schwarz. How do you prove that? – SimCity Sep 24 '12 at 14:42
Whoops :) Forgot to do that. Here you go: By the Lagrange form of the remainder in the Taylor expansion of $e^z - 1$, we have $|e^z - 1| \leq |w|$ for some $w \in D$. In particular, we have $|e^z - 1| \leq 1 \leq e - 1$, giving $|f(z)| \leq 1$. – Yoni Rozenshein Sep 24 '12 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.