# Taylor approximation of complex exponential function

I'm reading a paper which states the following error estimate for the Taylor approximation of the complex-valued exponential function.

For $$z \in \mathbb{C}$$ and $$d > 0$$, $$\left| e^z - \sum_{j=0}^{d-1} \frac{z^j}{j!} \right| \le O(1)\frac{|z|^d}{d!} \cdot \max\{1, e^{\Re(z)}\}.$$

It mentions that it follows from the Taylor series of the exponential function but I don't see how to derive this bound.

It would be helpful if someone could show me how to obtain this bound or provide a reference. Thanks.

I don't know how this follows from the Taylor expansion but I can give different proof. Let us first prove this when $$d=1$$. Consider $$\frac {e^{z}-1} {ze^{x}}$$ where $$x=Re z$$. Suppose $$|z|>1$$ and $$x>0$$. This ratio is then bounded because $$|e^{z}-1|\leq e^{x}+1$$ and $$\frac {1+e^{x}} {e^{x}}\leq 2$$. It is easier to prove boundedness of $$\frac {e^{z}-1} {z\max \{1, e^{x}\}}$$ when $$x<0$$ and I will leave that part to you. We have proved that $$|e^{z}-1| \leq C |z| \max \{1, e^{x}\}$$ for some $$C$$ whenever $$|z|>1$$ and the case $$|z| \leq 1$$ is simpler. Now observe that $$e^{z}-\sum\limits_{k=0}^{d-1} \frac {z^{k}} {k!}$$ is obtained by repeatedly integrating $$e^{z}-1$$ along the line segment from $$0$$ to $$z$$. For example, $$\int_0^{z} (e^{\zeta}-1)d\zeta=e^{z}-1-\frac z {1!}$$, etc. So the required inequality can be derived by induction on $$d$$.