# Prove this inequality for $|z| \leq 1$

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!

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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}

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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 $z$, $|\mathrm e^z-1|\leqslant|z|\cdot(\mathrm e^{|z|}-1)$.

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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.

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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