Show that $|e^{i\epsilon e^{i\theta}}-1| < 2\epsilon$ for all $\theta$, $0<\theta \le \pi$ for $\epsilon \in (0,1/2)$.

Question

I have this inequality from a proof in complex analysis but I cannot prove it.

For $0 < \epsilon < 1/2$ how can we show that $$|e^{i\epsilon e^{i\theta}}-1| < 2\epsilon$$ for all $\theta$, $0<\theta \le \pi$ expanding $e^{i\epsilon e^{i\theta}}$ in a power series?

Answer 1

$$\begin{align}\left\lvert e^{i\epsilon e^{i\theta}}-1\right\rvert&=\left\lvert \sum_{n=0}^\infty{1\over n!}\left(i\epsilon\right)^ne^{in\theta}-1\right\rvert\\ &=\left\lvert \sum_{n=1}^\infty{1\over n!}\left(i\epsilon\right)^ne^{in\theta}\right\rvert\\ &\leq \sum_{n=1}^\infty\left\lvert{1\over n!}\left(i\epsilon\right)^ne^{in\theta}\right\rvert\\ &=\sum_{n=1}^\infty{\epsilon^n\over n!}\\ &=e^\epsilon-1 \end{align}$$

so we just have to show that $e^x-1<2x,$ for $0<\epsilon<\frac12.$

Write $f(x)=e^x-2x-1.$ Then $f(0)=0$ and $$f'(x)= e^x-2,$$ so that $f'(x)<0$ when $0<x<\ln{2}\approx.6931$ Therefore, $f$ negative when $0<x<\frac12$.$

Answer 2

(This isn't strictly an answer, just a long comment, because it doesn't use the indicated power series expansion.)

Writing $x = \cos\theta$, $y = \sin\theta$, $z = e^{i\epsilon e^{i\theta}}$, and $r = |z|$, we have \begin{gather*} |x| \leqslant 1, \text{ and } 0 \leqslant y \leqslant 1; \\ z = e^{i\epsilon(x + iy)} = e^{-\epsilon y}e^{i\epsilon x}; \\ r = e^{-\epsilon y}, \text{ therefore } 1 \geqslant r \geqslant 1 - \epsilon y \geqslant 1 - \epsilon; \\ |z - r| = r|e^{i\epsilon x} - 1| = 2r|\sin(\epsilon x/2)| \leqslant r\epsilon|x| \leqslant \epsilon; \\ \text{therefore } |z - 1| \leqslant |z - r| + |r - 1| \leqslant 2\epsilon. \end{gather*}

There is only equality on the last line if there is equality on the line before; but this requires both $x = 0$ and $x = 1$, which is impossible; hence, $|z - 1| < 2\epsilon$.

(I don't seem to have used the assumption that $\epsilon < 1/2$, so perhaps there's a mistake.)