Is there a complex variable proof of $|e^z-1| \ge 1-1/e$ on the unit circle $|z|=1$?

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This could be done using Calculus by parametrizing the circle $z=e^{i\theta}$ and finding the critical points of $|e^{e^{i\theta}}-1|^2$ , however this approach is quite cumbersome. Is there a simple complex variables proof of the inequality?

This is an edited but much more satisfactory (imho) answer that uses just complex analysis to prove the result. The price is that we require a nontrivial theorem in geometric function theory, namely that if $F:\mathbb D \to \mathbb C$ is an univalent holomorphic function with convex image then $\mathcal L(F)=\frac{1}{z}\int_0^zF(u)du$ is also such, so in particular it is univalent. This has been proven by Libera (pdf at the link)

Hence with $F(z)=e^z$ which is clearly univalent in the unit disc and for which $\Re \frac{zF''}{F'}+1 =\Re z +1 >0$ satisfies the convexity condition, it follows that $g(z)=\mathcal L(F)=\frac{e^z-1}{z}$ is convex which is not immediately easy to prove. This immediately implies that $g(\partial \mathbb D)$ intersects the real line in at most two points which are $g(\pm 1)$ .

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