Remainder term in Taylor series for the complex exponential map The statement is the following: Let $P_n(z) = 1 + z + \frac{z^2}{2!} + ... + \frac{z^n}{n!}$ be the $n$ first terms of the Taylor series for $e^z$ centered at the origin. Then, prove that $\forall n \in \mathbb{N}$, and $\forall z \in \mathbb{C} : Re(z)< 0$, this inequality holds: 
$|e^z - P_n(z)|<|z|^{n+1}$
I am trying to use the integral formula for the remainder: 
https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_in_complex_analysis
And then the fact that in the half plane with $Re(z) < 0$, the exponential map has modulus less than one. But this bound is maybe too coarse, because I cannot reach the conclusion. Any help will be much appreciated. 
 A: Use $z\int_0^{1} \{e^{tz} -P_{n-1}(tz)\}dt =e^{z} -P_{n}(z)$. A simple induction argument gives the desired inequality. [To begin with $|\int_0^{1}e^{tz}dt| \leq 1$ for $\Re (z) <0$ so $|e^{z} -1| \leq |z|$
A: Solution: So I think I finally cracked and it goes like this. I used the result from https://math.stackexchange.com/a/2719893/527701 which was very useful. Let
$$P_n(z) = \sum_{k=0}^{n} \frac{z^k}{k!}$$ be the truncation taylor approximation of $exp(z)$ and we are interested in calculating the truncation error i.e. $|e^z-P_n(z)|$. We cannot make use of taylor theorem for reals i.e. the remainder term of the form $R_n(z) = \frac{f^{(n+1)}(\xi)z^{n+1}}{(n+1)!} $ where $\xi$ is between 0 and z because this theorem doesn't hold for complex (as far as I know after my research). This is because in complex analysis, Analytical function are equivalent to Holomorphic function. I.e. if a function is analytic, then in-fact it is infinite time complex differentiable. So I wasn't sure if I can say that there is a radius $r$ such that for $\xi\in B(0,r)$, the above bound holds where $B(0,r)$ is an open ball centred at zero with radius $r$. Moving onto proving this statement.
Proof by Induction:

*

*Take $n=0$ and the convention that $P_{-1}(z) = 0$, then using the result that
$$e^z - P_n(z) = z\int_0^1 e^{tz} - P_{n-1}(tz) \ dt$$ we can show that $|e^z-1| \leq |z|\int_0^1|e^{tz}|dt = |z|\int_0^1|e^{tRe(z)}|dt \leq |z|e^{Re(z)}\int_0^1dt = |z|e^{Re(z)} \leq |z| $ using the fact that exponential is an increasing function and $Re(z) <0$.


*Assume this holds for all $n$ i.e. $|e^z - P_{n-1}(z)| \leq \frac{|z|^n}{n!}$ and let us now consider $|e^z - P_{n}(z)|$.
$$|e^z - P_n(z)| \leq |z|\int_0^1 \frac{t^{n}|z|^n}{n!}dt = \frac{|z|^{n+1}}{(n+1)!} \leq |z|^{n+1}$$ since $n\in\mathbb{N}$. $\quad \square$
I hope this proof is useful for others. Also please let me know if you spot any mistakes or inconsistency.
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