Showing that $|e^z - 1|\leq |z|$ Assume that $z$ is complex with $\text{Re}(z) \leq 0$.

I'm trying to show that $$ |e^z - 1| \leq |z|$$
and, similarly,
$$ |e^z - z - 1| \leq |z|^2/2$$ holds.

The formulation of the exercise kind of hints to the series expansion of $e^z$, namely,
$$ e^z - 1 = z + \frac{z^2}{2} + \frac{z^3}{6} + o(z^4)$$
and similarly with $e^z - z - 1$. The $o(z)$ terms could probably be discarded applying triangle inequality, i.e.,
$$ |e^z - 1| = |z + o(z^2)| \leq |z|+|o(z^2)|$$
and
$$ \left|e^z -z -1 \right| = \left|\frac{z^2}{2} + o(z^3) \right| \leq \left|\frac{z^2}{2}\right|+|o(z^3)|$$
but I don't see why did we needed the negative real part here. Since $z$ is complex and I'm not used to that, I feel like I'm missing something (i.e., thinking about it, not entirely sure that $|z^2|=|z|^2$ in this case). Where am I wrong?
 A: Let $z=a+bi$ with $a\le 0$. Then,
$$
|e^z-1|^2=e^{2a}-2e^a\cos b+1=(e^a-1)^2+2e^a(1-\cos b)
$$
$$
\le (e^a-1)^2+e^ab^2\le (e^a-1)^2+b^2
$$
(I used the inequalities $e^a\le 1$ for $a\le 0$ and $1-\cos b\le b^2/2$.
We end by observing that $(e^a-1)^2\le a^2$ for $a\le 0$.
A: @ Robert, you are right. You can proceed by induction. Assuming that $|e^z-\frac{z^n}{n!}-\dots -z-1|\le \frac{|z|^{n+1}}{(n+1)!}$ , you can prove that it also holds with $n$ replaced by $n+1$ exactly in the same way you proceed to prove the inequality for $n=1$ from the case $n=0$. 
This works just because the derivative of $e^z-\frac{z^{n+1}}{(n+1)!}-\dots -z-1$ is precisely $e^z-\frac{z^{n}}{n!}-\dots -z-1$. Using the inductive hypothesis you arrive at the upper bound  $\int\frac{|\gamma|^{n+1}(t)}{(n+1)!}|\gamma'(t)|dt$ which is equal to $\frac{|z|^{n+2}}{(n+2)!}$.
A: Setting
$z = x + iy \tag 1$
with 
$x \le 0, \tag 2$
we find that
$\vert e^z \vert = \vert e^{x + iy} \vert = \vert e^x \vert \vert e^{iy} \vert = e^x \le 1; \tag 3$
if $\gamma(t)$ is the  path joining $z$ and $w$ in the half-plane $\{u \in \Bbb C, \Re(u) \le 0 \}$ given by
$\gamma(t) = tw + (1 - t)z = z + t(w - z), \tag 4$
then since $e^z$ is the primitive of itself, that is, $(e^z)' = e^z$, we may write
$e^w - e^z = \displaystyle \int_0^1 (\exp(\gamma(t))' \; dt$
$= \displaystyle \int_0^1 \exp(\gamma(t)) \gamma'(t) \; dt = \int_0^1 \exp(\gamma(t)) (w - z) \; dt; \tag 5$
thus, by virtue of (3), 
$\vert e^w - e^z \vert \le \displaystyle \int_0^1 \vert \exp(\gamma(t)) \vert \vert w - z \vert \; dt \le \vert w - z \vert; \tag 6$
set 
$w = 0 \tag 7$
and find
$\vert e^z - 1 \vert = \vert e^z - e^0 \vert \le \vert z \vert, \tag 8$
which is the first desired inequality; next, note that
$(e^z - z - 1)' = e^z - 1; \tag 9$
$e^z - z - 1 = (e^z - z - 1) - (e^0 - 0 - 1) = \displaystyle \int_0^1 (\exp(\gamma(t) - 1)) \gamma'(t) \; dt, \tag{10}$
whence, using (8),
$\vert e^z - z - 1 \vert \le \displaystyle \int_0^1 \vert \exp(\gamma(t) - 1 \vert \vert \gamma'(t) \vert \; dt$
$\le \displaystyle \int_0^1 \vert \gamma(t) \vert \vert \gamma'(t) \vert \; dt = \int_0^1 \vert (1 - t)z \vert \vert z \vert \; dt = \vert z \vert^2 \int_0^1 (1 - t) \; dt  = \dfrac{\vert z \vert^2}{2}, \tag{11}$
the second inequality whose proof was sought.  $OE\Delta$.
Nota Bene:  I cannot help but wonder at this point if these results may not be extended to show that
$\left \vert e^z - \displaystyle \sum_0^n \dfrac{z^k}{k!} \right \vert \le \dfrac{\vert z \vert^{n + 1}}{(n + 1)!}, \tag{12}$
or some similar inequality; I suspect this is so but have not yet a complete proof, merely ideas; for example, we may be able to build (11) for larger $n$ by building upon established inequalities for lesser $n$, in a manner analogous to the way we have arrived at (11) based upon (8), etc. etc. etc. End of Note.
A: Note that $$|e^{z}-1|=|\int_{0}^{z}{e^{s}ds}|\leq \int_{0}^{z}{|e^{s}||ds|}= \int_{0}^{z}{e^{\Re(s)}|ds|}\leq \int_{0}^{z}|ds|=|z|$$
Since $\Re(s)\leq 0.$
