Bound on $|e^z - 1|$ I'm trying to show that $|e^z - 1| < e - 1$ when $|z| < 1$.  The hint is to consider $e^z - 1 = \int_0^z e^w dw$.  Thank you!
 A: Let $\gamma$ be the straight line connecting $0$ to $z$. Observe that
$$
e^z - 1 = \int_\gamma e^w \, dw = \int^1_0 z \cdot e^{tz} \, dt = z \cdot \int^1_0 e^{tz}\,dt.
$$
We then have
$$
|e^z-1| = \left| \int_\gamma e^w \, dw \right| = |z| \cdot \left| \int^1_0 e^{tz}\,dt \right| \leq \int^1_0 |e^{tz}|\,dt = \int^1_0 e^{\Re(tz)}\,dt,
$$
where $\Re(tz)$ is the real part of $tz$. We have $\Re(tz) < t$ (why?), so that
$$
\int^1_0 e^{\Re(tz)}\,dz < \int^1_0 e^t = e - 1\,dt.
$$
A: Consider $\vert e^z -1 \vert = \vert \int_0^z e^w dw \vert$.  Since $e^w$ is holomorphic for $\vert w \vert < 1$, the path we choose to evaluate $\int_0^z e^w dw$ may be freely chosen within the open unit disk $\{z: \vert z \vert<1 \}$; for $z = re^{i \theta}$, we choose the path $w(\rho) = \rho e^{i\theta}$ with $\theta$ constant and $0 \le \rho \le r <1$.  Then $dw = e^{i \theta}d\rho$, and $e^{w(\rho)} = e^{\rho e^{i\theta}} = e^{\rho \cos \theta +i \rho \sin \theta} = e^{\rho \cos \theta}e^{i \rho \sin \theta}$.  Thus 
$\vert \int_0^z e^w dw \vert = \vert \int_0^r e^{w(\rho)} e^{i\theta} d\rho \vert  =  \vert \int_0^r e^{\rho \cos \theta} e^{i \rho \sin \theta} e^{i\theta} d\rho \vert \le \int_0^r \vert e^{\rho \cos \theta} e^{i \rho \sin \theta} e^{i \theta} \vert d\rho$
$= \int_0^r e^{\rho \cos \theta} d\rho \le \int_0^r e^\rho d\rho  = e^r -1 < e - 1 \tag{1}$
since $r < 1$.  We have used the facts that $\vert e^{i \rho \sin \theta} \vert = \vert e^{i \theta} \vert = 1$ and $e^{\rho \cos \theta} \le e^\rho$ (since $\cos \theta \le 1$) in forming this estimate.  QED.
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
A: Let $z = a + bi$ with $a^2 + b^2 < 1$. So $|e^z - 1|^2 = |e^a\cdot e^{ib} - 1|^2 = |e^a\cdot cosb - 1 +  i\cdot e^a \cdot sinb|^2 = (e^a\cdot cosb - 1)^2 + e^{2a}\cdot sin^2b = e^{2a} - 2\cdot e^a \cdot cosb + 1 = f(a,b)$.  
Taking partial derivatives of $f$ with respect to $a$ and $b$ and set them equal to $0$ we have:
$\dfrac{\partial f}{\partial a} = 2\cdot e^{2a} - 2\cdot e^a\cdot cosb = 0$
$\dfrac{\partial f}{\partial b} = 2\cdot e^a\cdot sinb = 0$
Solve this system we have: $b = 0$ and $a = 0$. And this gives $f_{min} = f(0,0) = 0$.
For finding $f_{max}(a,b)$, let $a \to 0^+$ and $b \to 1^-$, and let $a \to 1^-$ and $b \to 0+$we have: $f_{max}(a,b) < max\{f(0,1), f(1,0)\} = max\{0.919395,(e - 1)^2\}$. Thus $|e^z - 1| < \sqrt{max\{0.919395,(e - 1)^2\}} = e - 1$
