Prove that for any $z \in \mathbb{C}$, such that $\mathcal{R(z)}<0$, the inequality $|1-e^z|<|z|$ is true.
I am thinking to use Maximal Modulus principle to prove this theorem. However, I am not sure about my proof and I need help to verify my approach.
Define $g(z)=\frac{e^z-1}{z}$ which is analytic in left half plane. Applying the maximal modulus principal the maximum of $g(z)$ should occur on the boundary of left half plane.
Let $R$ large enough, for any $z\in D^{-}(0,R)$ (the left halpf palne) , we have
$$|g(z)|=\left|\frac{e^z-1}{z}\right|\leq \max{\left\{\left|\frac{e^z-1}{z}\right|:z\in D^{-}(0,R)\right\}}$$ $$=\left\{\left|\frac{e^z-1}{z}\right|:z\in C^{-}(0,R)\right\}\leq\frac{2}{R}$$
Note that $\mathcal{R}(z)<0$, $|e^z|=|e^{\mathcal{R}(z)}|<1$.
Now since $R$ can be large neough so $\frac{2}{R}<1$. It proves the theorem.
There is a very simple proof of this: $|\frac {1-e^{z}} {z}|=|\int_0^{1} e^{tz}\, dt| \leq \int_0^{1}|e^{tz}|\, dt =\int_0^{1} e^{-t\Re z} \, dt < 1$