Prove Complex analysis inequality

Question

Prove $ | \mathbb{e^z}-1| \le |z|$ if $Re(z) < 0$.

Any hints on how to start?

Answer 1

Another, perhaps more complex-analytic, way: Let $\gamma$ be the line segment from $0$ to $z$. Then $$ e^z - 1 = \int_\gamma e^w\,dw, $$ so $$ |e^z - 1| \le \left| \int_\gamma e^w\,dw \right| \le \max_{w\in\gamma} |e^w| \cdot \ell(\gamma) \le 1 \cdot |z| $$ since $|e^w| \le 1$ when $\operatorname{Re}(w) \le 0$ (which is true on $\gamma$ since $\operatorname{Re}(z) \le 0$).

Answer 2

HINT:

$$|e^z-1|^2=(e^x-1)^2+4e^x\sin^2(y/2)$$

Then, use

$$|\sin(y)|\le |y|$$

and for $x<0$

$$|e^x-1|\le |x|$$

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Let $z=x+iy$. Then, $e^{z}=e^{x+iy}=e^x\cos(y)+ie^x\sin(y)$. Therefore, we can write $$\begin{align}\left|e^z -1\right|^2&=(e^x\cos(y)-1)^2+e^{2x}\sin^2(y)\\\\&=e^{2x}-2e^x\cos(y)+1\\\\&=(e^x-1)^2+2e^2(1-\cos(y))\\\\&=(e^x-1)^2+4e^x\sin^2(y/2) \tag 1\end{align}$$Recall (from basic geometry) that the sine function satisfies the inequality $$|\sin(x)|\le x \tag 2$$In addition, I showed in THIS ANSWER and THIS ONE using only (i) the limit definition of the exponential function and (ii) Bernoulli's Inequality that the exponential function satisfies the inequality $$e^x-1 \ge x$$Now, for $x<0$, we have $$|e^x-1|\le |x| \tag 3$$Using the inequalities in $(2)$ and $(3)$ in $(1)$ yields $$\begin{align}|e^z-1|^2&\le x^2+4e^x(y/2)^2\\\\&=x^2+e^xy^2\\\\&<x^2+y^2\\\\&=|z|^2 \tag 4\end{align}$$Taking square roots on both sides of $(4)$ gives the inequality for $x<0$ $$|e^z-1|\le |z|$$And we are done!