on the size of $e^{i \theta } - 1 - i \theta$? For any $\theta \in \mathbb{R}$, it is well known
$$
|e^{i\theta}| = 1. 
$$
It is follows that
$$
|e^{i \theta} - 1| \leq 2
$$
even though the Taylor series look like
$$
e^{i \theta} - 1 = i \theta + \frac{(i \theta)^2}{2!} + ...
$$
Thus even if $\theta$ is very large, things cancel in this case.
Does this continue to the next term is my question?
In other words, does there exist a constant $C > 0$ such that
$$
|e^{i \theta} - 1 - i \theta| \leq C
$$
for all $\theta \in \mathbb{R}$.
I don't think such $C$ exists but I couldn't find a definitive way to convince myself.... any suggestion appreciated
 A: No, because $$|\theta| = |i\theta| \le |e^{i\theta} - 1 - i\theta| + |1 - e^{i\theta}| \le |e^{i\theta} - 1 - i\theta| + 2.$$ If the right hand side were to be bounded, so would be the left hand side.
A: Here another way: if $z_1=i\theta$ and $z_2=1+i\theta$ then
$$
\begin{array}{l}
 \left| {z_1  - z_2 } \right| \ge \left| {\left| {z_1 } \right| - \left| {z_2 } \right|} \right| = \left| {\left| {e^{i\theta } } \right| - \left| {1 + i\theta } \right|} \right| =  \\ 
  \\ 
  = \left| {1 - \sqrt {1 + \theta ^2 } } \right| = \sqrt {1 + \theta ^2 }  - 1 \\ 
 \end{array}
$$
Since $$
\sqrt {1 + \theta ^2 }  - 1
$$
in not bounded from above you get that no $C>0$ which bound the absolute value of your function can exists.
A: You should think of this situation as wrapping a string ($w=i\theta$) around the unit circle. For small $\theta$, the points on the string are close to the points they wind to; for larger, the point on the string has to move rather farther to get to its assigned point on the circle.
