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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?

Let $z=a+bi$ with $ale 0$. Then,
$$
|e^z-1|^2=e^{2a}-2e^acos b+1=(e^a-1)^2+2e^a(1-cos b)
$$
$$
le (e^a-1)^2+e^ab^2le (e^a-1)^2+b^2
$$
(I used the inequalities $e^ale 1$ for $ale 0$ and $1-cos ble b^2/2$.

We end by observing that $(e^a-1)^2le a^2$ for $ale 0$.

@ 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 $intfrac{|gamma|^{n+1}(t)}{(n+1)!}|gamma'(t)|dt$ which is equal to $frac{|z|^{n+2}}{(n+2)!}$.

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. $OEDelta$.

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.

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 $Re(s)leq 0.$